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Some Applications of Trigonometry: 5 marks Questions (CBSE Class 10)

73 different 5 marks questions on Some Applications of Trigonometry from CBSE Class 10 Maths board exams 2022–2026, newest first.

1 mark (30)2 marks (3)3 marks (20)4 marks (25)5 marks (73)

Q345 marksLong AnswerSome Applications of Trigonometry CBSE 2026 · Basic 430/4/1

The angle of elevation of the top of a building from the foot of the tower is $3 0^{\circ}$ and the angle of elevation of the top of the tower from the foot of the building is $6 0^{\circ}$ . If the tower is 60 m high, find the height of the building and distance between the building and the tower. (Use $3$ = 1.73)

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Answer: Height of building = 20 m; distance =

203

m = 34.6 m

Let the distance between building and tower be d and the height of the building be h.
From the foot of the building: $tan 6 0^{\circ} = \frac{60}{d} \Rightarrow d = \frac{60}{3} = 203$ m.
From the foot of the tower: $tan 3 0^{\circ} = \frac{h}{d} \Rightarrow h = \frac{20 3}{3} = 20$ m.
d = $20 \times 1.73 = 34.6$ m.

Q355 marksLong AnswerSome Applications of Trigonometry CBSE 2026 · Basic 430/4/2

A statue, 2 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is $6 0^{\circ}$ and from the same point the angle of elevation of the bottom of the statue is $4 5^{\circ}$ . Find the height of the pedestal and its distance from the point of observation on the ground. (use $3$ = 1.73)

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Answer: Height of pedestal =

(3 + 1)

m = 2.73 m; distance = 2.73 m

Let the pedestal height be h m and its distance from the point be d m.
$tan 4 5^{\circ} = \frac{h}{d} \Rightarrow d = h$ .
$tan 6 0^{\circ} = \frac{h + 2}{d} \Rightarrow 3 h = h + 2$ .
$h = \frac{2}{3 - 1} = \frac{2 ( 3 + 1 )}{2} = 3 + 1 = 2.73$ m.
Distance d = h = 2.73 m.

Q325 marksLong AnswerSome Applications of Trigonometry CBSE 2026 · Basic 430/4/3

A straight road leads to foot of a tower whose shadow is found to be 40 m longer when sun's altitude is $3 0^{\circ}$ than when it is $6 0^{\circ}$ . Find the height of the tower and length of shadow in both the situations. (Use $3$ = 1.73)

Show answer & solution

Answer: Height =

203

m = 34.6 m; shadow = 20 m at

6 0^{\circ}

and 60 m at

3 0^{\circ}

Let the height of the tower be h m.
Shadow at $6 0^{\circ}$ : $\frac{h}{t a n 6 0 ^{\circ}} = \frac{h}{3}$ ; shadow at $3 0^{\circ}$ : $\frac{h}{t a n 3 0 ^{\circ}} = 3 h$ .
$3 h - \frac{h}{3} = 40 \Rightarrow \frac{2 h}{3} = 40 \Rightarrow h = 203 = 34.6$ m.
Shadow at $6 0^{\circ}$ = $\frac{20 3}{3} = 20$ m.
Shadow at $3 0^{\circ}$ = $3 \times 203 = 60$ m.

Q335 marksLong AnswerSome Applications of Trigonometry CBSE 2026 · Basic 430/5/1

Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are $6 0^{\circ}$ and $3 0^{\circ}$ respectively. Find the height of the poles and the distance of the point from the poles.

Show answer & solution

Answer: Height

= 203

m (

\approx 34.64

m); the point is 20 m from one pole and 60 m from the other.

Let each pole have height h m and let the point be x m from the pole whose top is seen at 60°; it is (80 − x) m from the other pole.
$tan 6 0^{\circ} = \frac{h}{x} \Rightarrow h = 3 x$
$tan 3 0^{\circ} = \frac{h}{80 - x} \Rightarrow h = \frac{80 - x}{3}$
$3 x = \frac{80 - x}{3} \Rightarrow 3 x = 80 - x \Rightarrow x = 20$
$h = 203 \approx 34.64$ m; distances are 20 m and 60 m.

Also asked in: 2026 Basic 430/5/2, 2026 Basic 430/5/3

Q355 marksLong AnswerSome Applications of Trigonometry CBSE 2026 · Standard 30/3/1

A boy standing on a horizontal plane is flying a kite with a string of length 60 m, at an angle of elevation of $3 0^{\circ}$ . Another boy standing on the roof of a 20 m high building, finds the angle of elevation of same kite to be $4 5^{\circ}$ . If both the boys are on opposite sides of the kite, find the distance of the first boy from the base of the building. Also, find the height of the kite from the ground. (Use $3$ = 1.73)

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Answer: Distance = 61.9 m; height of kite = 30 m

Let K be the kite and M the point on the ground directly below it
First boy at A: AK $= 60$ m, $∠ K A M = 3 0^{\circ}$
KM $= 60 sin 3 0^{\circ} = 30$ m (height of kite)
AM $= 60 cos 3 0^{\circ} = 303 = 51.9$ m
Second boy at roof point R, 20 m high: height of kite above roof $= 30 - 20 = 10$ m
$tan 4 5^{\circ} = \frac{10}{horizontal distance}$ , so horizontal distance from building to M $= 10$ m
Boys are on opposite sides, so distance of first boy from base of building $= 51.9 + 10 = 61.9$ m

Q325 marksLong AnswerSome Applications of Trigonometry CBSE 2026 · Standard 30/3/2

A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff. From a point on the ground 30 m away from the tower, wires are attached to the top and bottom of the flagstaff making angles of elevation $6 0^{\circ}$ and $3 0^{\circ}$ respectively. Find the height of the tower and lengths of the wires attached. (Take $3$ = 1.73)

Show answer & solution

Answer: Height of tower

= 103 = 17.3

m; wires:

203 = 34.6

m (to bottom of flagstaff) and 60 m (to top)

Let the point be A, foot of tower B, top of tower (bottom of flagstaff) C, top of flagstaff D; AB = 30 m
In right $△ A B C$ : $tan 3 0^{\circ} = \frac{B C}{30}$ , so BC $= \frac{30}{3} = 103 = 17.3$ m
Wire AC $= \frac{30}{c o s 3 0 ^{\circ}} = \frac{60}{3} = 203 = 34.6$ m
In right $△ A B D$ : $cos 6 0^{\circ} = \frac{30}{A D}$ , so wire AD $= 60$ m
(Height of flagstaff $= 303 - 103 = 203 = 34.6$ m)

Q345 marksLong AnswerSome Applications of Trigonometry CBSE 2026 · Standard 30/3/3

The angle of elevation of the top of a building from a point A, on the ground, is $3 0^{\circ}$ . On moving a distance of 24 m towards its base to the point B, the angle of elevation changes to $6 0^{\circ}$ . Find the height of the building and distance of point A from the base of the building. (Take $3$ = 1.73)

Show answer & solution

Answer: Height

= 123 = 20.76

m; distance of A from base

= 36

Let the building be CD (C the base), height $h$ , and BC $= x$
From B: $tan 6 0^{\circ} = \frac{h}{x}$ , so $h = 3 x$
From A: $tan 3 0^{\circ} = \frac{h}{x + 24}$ , so $x + 24 = 3 h = 3 x$
$x = 12$ m, $h = 123 = 12 \times 1.73 = 20.76$ m
Distance of A from the base $= 12 + 24 = 36$ m

Q34 (OR) (OR)5 marksLong AnswerSome Applications of Trigonometry CBSE 2026 · Standard 30/3/3

A tower stands vertically on the ground. A man standing at the top of the tower observes his friend at an angle of depression of $3 0^{\circ}$ , who is approaching the foot of the tower with a uniform speed. 30 seconds later, the angle of depression changes to $6 0^{\circ}$ . Find the time taken by his friend to reach the foot of the tower from this point.

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Answer: 15 seconds

Let the height of the tower be $h$
At $3 0^{\circ}$ depression: distance from foot $= \frac{h}{t a n 3 0 ^{\circ}} = 3 h$
At $6 0^{\circ}$ depression: distance from foot $= \frac{h}{t a n 6 0 ^{\circ}} = \frac{h}{3}$
Distance covered in 30 s $= 3 h - \frac{h}{3} = \frac{2 h}{3}$
Remaining distance $= \frac{h}{3}$ , which is half of that
At the same speed, time $= \frac{30}{2} = 15$ seconds

Q345 marksLong AnswerSome Applications of Trigonometry CBSE 2026 · Standard 30/4/1

A kite is flying at a height of 60 m above the ground level. Ravi, standing at the roof of the house is holding the string straight and observes the angle of elevation of kite as $3 0^{\circ}$ . From the bottom of the same building, the angle of elevation of kite is $4 5^{\circ}$ . Find the length of the string and height of roof from the ground. (Use $3 = 1.73$ )

Show answer & solution

Answer: Length of string = 69.2 m; height of roof = 25.4 m

Let the building be AB (B at the ground, A at the roof) of height h, the kite at K, 60 m above the ground, and d the horizontal distance of the kite from the building.
From B: $tan 4 5^{\circ} = \frac{60}{d} \Rightarrow d = 60$ m
From A: $tan 3 0^{\circ} = \frac{60 - h}{d} \Rightarrow 60 - h = \frac{60}{3} = 203 = 34.6$
$h = 60 - 34.6 = 25.4$ m
String $A K = \frac{d}{c o s 3 0 ^{\circ}} = \frac{60 \times 2}{3} = 403 = 69.2$ m

Also asked in: 2026 Standard 30/4/2, 2026 Standard 30/4/3

Q355 marksLong AnswerSome Applications of Trigonometry CBSE 2025 · Basic 430/4/1

From a point on the ground, the angle of elevation of the top of a pedestal is $3 0^{\circ}$ and that of the top of the flagstaff fixed on the pedestal is $6 0^{\circ}$ . If the length of the flagstaff is 5 m, then find the height of the pedestal and its distance from the point of observation on ground. (Use $3 = 1.73$ )

Show answer & solution

Answer: Height of pedestal = 2.5 m; distance =

2.5 3 \approx 4.33

Let the pedestal height be $h$ m and its distance from the point be $x$ m.
$tan 3 0^{\circ} = \frac{h}{x}$ , so $x = h 3$ .
$tan 6 0^{\circ} = \frac{h + 5}{x}$ , so $h + 5 = 3 x = 3 h$ .
$2 h = 5$ , so $h = 2.5$ m.
$x = 2.5 3 = 2.5 \times 1.73 = 4.325 \approx 4.33$ m.

Q335 marksLong AnswerSome Applications of Trigonometry CBSE 2025 · Basic 430/4/2

As observed from the top of a 70 m high lighthouse from the sea level, the angles of depression of two ships are $3 0^{\circ}$ and $4 5^{\circ}$ . If one ship is exactly behind the other on the same sides of the lighthouse, find the distance between the two ships. (Use $3 = 1.73$ )

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Answer:

70 (3 - 1) = 51.1

Let AB = 70 m be the lighthouse with top A, and C, D the ships (D farther), on a line through B.
Angle of elevation of A from each ship equals the corresponding angle of depression.
In right $△ A B C$ : $tan 4 5^{\circ} = \frac{70}{B C}$ , so BC = 70 m.
In right $△ A B D$ : $tan 3 0^{\circ} = \frac{70}{B D}$ , so $B D = 703$ m.
Distance between ships $= B D - B C = 70 (3 - 1) = 70 \times 0.73 = 51.1$ m.

Q325 marksLong AnswerSome Applications of Trigonometry CBSE 2025 · Basic 430/4/3

The angle of elevation of the top of a tower, 300 m high, from a point on the ground is observed as $3 0^{\circ}$ . At an instant a hot air balloon passes vertically above the tower and at that instant its angle of elevation from same point on the ground is $6 0^{\circ}$ . Find height of the balloon from the ground and distance of tower from point of observation. (Use $3 = 1.73$ )

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Answer: Height of balloon = 900 m; distance of tower =

3003 = 519

Let the tower be AB = 300 m, the point of observation P, and PB = $x$ m.
$tan 3 0^{\circ} = \frac{300}{x}$ , so $x = 3003 = 300 \times 1.73 = 519$ m.
The balloon C is vertically above B, so $tan 6 0^{\circ} = \frac{B C}{x}$ .
$B C = 3 \times 3003 = 900$ m.

Q345 marksLong AnswerSome Applications of Trigonometry CBSE 2025 · Basic 430/5/1

The angle of elevation of the top of a building from the foot of the tower is $3 0^{\circ}$ and the angle of elevation of the top of the tower from the foot of the building is $6 0^{\circ}$ . If the tower is 30 m high, find the height of the building and distance between the building and the tower. (Use $3 = 1.73$ )

Show answer & solution

Answer: Height of building = 10 m; distance =

103

m = 17.3 m

Let the distance between the building and the tower be $d$ and the height of the building be $h$ .
From the foot of the building: $tan 6 0^{\circ} = \frac{30}{d}$ , so $d = \frac{30}{3} = 103 = 17.3$ m.
From the foot of the tower: $tan 3 0^{\circ} = \frac{h}{d}$ , so $h = \frac{10 3}{3} = 10$ m.
Building height 10 m; distance 17.3 m.

Q325 marksLong AnswerSome Applications of Trigonometry CBSE 2025 · Basic 430/5/2

From a point on the ground, the angle of elevation of the top of a tree observed by a person is $6 0^{\circ}$ . When moved back by 28 m, in the same line, the angle of elevation from another point on ground becomes $3 0^{\circ}$ . Find the height of the tree and its distance from the initial point. (Use $3$ = 1.73)

Show answer & solution

Answer: Height =

143

m = 24.22 m; distance from the initial point = 14 m

Let the height of the tree be $h$ m and its distance from the initial point be $d$ m.
$tan 6 0^{\circ} = \frac{h}{d}$ , so $h = 3 d$ .
$tan 3 0^{\circ} = \frac{h}{d + 28}$ , so $h = \frac{d + 28}{3}$ .
$3 d = \frac{d + 28}{3}$ gives $3 d = d + 28$ , so $d = 14$ m.
$h = 143 = 14 \times 1.73 = 24.22$ m.

Q335 marksLong AnswerSome Applications of Trigonometry CBSE 2025 · Basic 430/5/3

A statue 3 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is $6 0^{\circ}$ and from the same point the angle of elevation of the top of the pedestal is $3 0^{\circ}$ . Find the height of the pedestal and its distance from the point of observation on ground. (Use $3$ = 1.73)

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Answer: Height of pedestal = 1.5 m; distance =

\frac{3 3}{2}

m = 2.595 m

Let the height of the pedestal be $h$ m and its distance from the point of observation be $d$ m.
$tan 3 0^{\circ} = \frac{h}{d}$ , so $d = 3 h$ .
$tan 6 0^{\circ} = \frac{h + 3}{d}$ , so $h + 3 = 3 d = 3 \times 3 h = 3 h$ .
$2 h = 3$ , so $h = 1.5$ m.
$d = 1.5 3 = 1.5 \times 1.73 = 2.595$ m.

Q345 marksLong AnswerSome Applications of Trigonometry CBSE 2025 · Basic 430/6/1

The angles of depression of the top and the foot of a 9 m tall building from the top of a multi-storeyed building are $3 0^{\circ}$ and $6 0^{\circ}$ respectively. Find the height of the multi-storeyed building and the distance between the two buildings. (Use $3 = 1.73$ )

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Answer: Height = 13.5 m; distance

= \frac{9 3}{2} \approx 7.79

m (7.785 m)

Let the multi-storeyed building be $h$ m high and the distance between the buildings be $d$ m.
Foot of the 9 m building (depression $6 0^{\circ}$ ): $tan 6 0^{\circ} = \frac{h}{d}$ , so $h = 3 d$ .
Top of the 9 m building (depression $3 0^{\circ}$ ): $tan 3 0^{\circ} = \frac{h - 9}{d}$ , so $h - 9 = \frac{d}{3}$ .
$3 d - \frac{d}{3} = 9$ , so $\frac{2 d}{3} = 9$ and $d = \frac{9 3}{2} = \frac{9 \times 1.73}{2} = 7.785$ m.
$h = 3 d = \frac{27}{2} = 13.5$ m.

Q325 marksLong AnswerSome Applications of Trigonometry CBSE 2025 · Basic 430/6/2

Two poles of equal heights are standing opposite each other on either side of the road which is 85 m wide. From a point between them on the road, the angles of elevation of the top of the poles are $6 0^{\circ}$ and $3 0^{\circ}$ respectively. Find the height of the poles and the distances of the point from the poles. (Use $3 = 1.73$ )

Show answer & solution

Answer: Height

= \frac{85 3}{4} \approx 36.76

m; distances 21.25 m and 63.75 m

Let the height be $h$ m and the point be $x$ m from the first pole (elevation $6 0^{\circ}$ ), so $(85 - x)$ m from the other.
$tan 6 0^{\circ} = \frac{h}{x}$ , so $h = 3 x$ .
$tan 3 0^{\circ} = \frac{h}{85 - x}$ , so $h = \frac{85 - x}{3}$ .
$3 x = \frac{85 - x}{3}$ gives $3 x = 85 - x$ , $x = 21.25$ m.
$h = 21.25 \times 1.73 = 36.7625 \approx 36.76$ m.
Distances from the poles: 21.25 m and 85 $-$ 21.25 = 63.75 m.

Q355 marksLong AnswerSome Applications of Trigonometry CBSE 2025 · Basic 430/6/3

A drone is flying at a height of h metres. At an instant it observes the angle of elevation of top of an industrial turbine as $6 0^{\circ}$ and angle of depression of foot of the turbine as $3 0^{\circ}$ . If height of turbine is 200 metres, find the value of h and the distance of drone from the turbine.
(Use $3 = 1.73$ )

Show answer & solution

Answer: h = 50 m; distance

= 503 = 86.5

Let the horizontal distance between the drone and the turbine be $d$ m.
Depression of foot: $tan 3 0^{\circ} = \frac{h}{d}$ , so $d = 3 h$ .
Elevation of top: $tan 6 0^{\circ} = \frac{200 - h}{d}$ , so $200 - h = 3 d = 3 h$ .
$4 h = 200$ , so $h = 50$ m.
$d = 503 = 50 \times 1.73 = 86.5$ m.

Q325 marksLong AnswerSome Applications of Trigonometry CBSE 2025 · Standard 30/2/1

Two ships are sailing in the sea on either side of a lighthouse. The angles of depression to the two ships as observed from the top of the lighthouse are $6 0^{\circ}$ and $4 5^{\circ}$ , respectively. If the distance between the ships is $100 (\frac{1 + 3}{3})$ m, then find the height of the lighthouse.

Show answer & solution

Answer: 100 m

Let the height be h m; the ships are at distances x and y from the foot on opposite sides
$tan 6 0^{\circ} = \frac{h}{x}$ , so $x = \frac{h}{3}$
$tan 4 5^{\circ} = \frac{h}{y}$ , so $y = h$
$x + y = h (\frac{1}{3} + 1) = h (\frac{1 + 3}{3})$
Equating with $100 (\frac{1 + 3}{3})$ : $h = 100$ m

Also asked in: 2025 Standard 30/2/3

Q32 (OR) (OR)5 marksLong AnswerSome Applications of Trigonometry CBSE 2025 · Standard 30/2/1

The angles of depression of the top and the bottom of an 8 m tall building from the top of another multistoried building are $3 0^{\circ}$ and $4 5^{\circ}$ , respectively. Find the height of the multistoried building and the distance between the two buildings.

Show answer & solution

Answer: Height

= 4 (3 + 3)

\approx 18.93

m; distance

= 4 (3 + 3)

\approx 18.93

Let the multistoried building have height H m and let the distance between the buildings be d m
Bottom of 8 m building: $tan 4 5^{\circ} = \frac{H}{d}$ , so $d = H$
Top of 8 m building: $tan 3 0^{\circ} = \frac{H - 8}{d}$ , so $H - 8 = \frac{H}{3}$
$H (3 - 1) = 83$ , so $H = \frac{8 3}{3 - 1} = \frac{8 3 ( 3 + 1 )}{2} = 12 + 43$
$H = d = 4 (3 + 3) \approx 18.93$ m

Also asked in: 2025 Standard 30/2/3

Q355 marksLong AnswerSome Applications of Trigonometry CBSE 2025 · Standard 30/2/2

The angle of elevation of an airborne helicopter from a point A on the ground is $4 5^{\circ}$ . After a flight of 15 seconds, the angle of elevation of the helicopter changes to $3 0^{\circ}$ . If the helicopter is flying at a constant height of 2000 m, find the speed of the helicopter. (Take $3 = 1.732$ )

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Answer: 97.6 m/s

Initial horizontal distance from A: $tan 4 5^{\circ} = \frac{2000}{x _{1}}$ , so $x_{1} = 2000$ m
Later: $tan 3 0^{\circ} = \frac{2000}{x _{2}}$ , so $x_{2} = 20003$ m
Distance flown $= 20003 - 2000 = 2000 (3 - 1) = 2000 \times 0.732 = 1464$ m
Speed $= \frac{1464}{15} = 97.6$ m/s

Q35 (OR) (OR)5 marksLong AnswerSome Applications of Trigonometry CBSE 2025 · Standard 30/2/2

A girl 1.5 m tall is standing at some distance from a 30 m high tower. The angle of elevation from her eye to the top of the tower increases from $3 0^{\circ}$ to $6 0^{\circ}$ as she walks towards the tower. Find the distance she walked towards the tower.

Show answer & solution

Answer:

193

\approx 32.91

Height of tower above her eye level $= 30 - 1.5 = 28.5$ m
Initial distance: $tan 3 0^{\circ} = \frac{28.5}{d _{1}}$ , so $d_{1} = 28.5 3$ m
Final distance: $tan 6 0^{\circ} = \frac{28.5}{d _{2}}$ , so $d_{2} = \frac{28.5}{3}$ m
Distance walked $= 28.5 3 - \frac{28.5}{3} = \frac{28.5 \times 2}{3} = \frac{57}{3} = 193 \approx 32.91$ m

Q335 marksLong AnswerSome Applications of Trigonometry CBSE 2024 · Basic 430/1/2

The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun's altitude is $3 0^{\circ}$ than when it was $6 0^{\circ}$ . Find the height of the tower and the length of original shadow. (use $3 = 1.73$ )

Show answer & solution

Answer: Height

= 203 = 34.6

m; original shadow

= 20

Let the height be $h$ and the original shadow (at $6 0^{\circ}$ ) be $x$ m.
$tan 6 0^{\circ} = \frac{h}{x}$ , so $h = 3 x$ .
$tan 3 0^{\circ} = \frac{h}{x + 40}$ , so $h = \frac{x + 40}{3}$ .
$3 x = \frac{x + 40}{3}$ gives $3 x = x + 40$ , so $x = 20$ m.
$h = 203 = 20 \times 1.73 = 34.6$ m.

Also asked in: 2024 Basic 430/1/3

Q33 (OR) (OR)5 marksLong AnswerSome Applications of Trigonometry CBSE 2024 · Basic 430/1/2

The angles of depression of the top and the bottom of an 8 m tall building from the top of a multi–storeyed building are $3 0^{\circ}$ and $4 5^{\circ}$ respectively. Find the height of the multi–storeyed building and the distance between the two buildings. (use $3 = 1.73$ )

Show answer & solution

Answer: Height

= 4 (3 + 3) = 18.92

m; distance

= 18.92

Let the multi-storeyed building have height $H$ m and let the distance between the buildings be $d$ m.
Angle of depression of the bottom is $4 5^{\circ}$ : $tan 4 5^{\circ} = \frac{H}{d}$ , so $d = H$ .
Angle of depression of the top is $3 0^{\circ}$ : $tan 3 0^{\circ} = \frac{H - 8}{d}$ , so $H - 8 = \frac{H}{3}$ .
$H (3 - 1) = 83$ , so $H = \frac{8 3}{3 - 1} = \frac{8 3 ( 3 + 1 )}{2} = 4 (3 + 3)$ .
$H = 4 \times 4.73 = 18.92$ m, and $d = H = 18.92$ m.

Also asked in: 2024 Basic 430/1/3

Q335 marksLong AnswerSome Applications of Trigonometry CBSE 2024 · Basic 430/2/1

A TV tower stands vertically on a bank of a canal. From a point on the other bank exactly opposite the tower, the angle of elevation of the top of the tower is $6 0^{\circ}$ . From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is $3 0^{\circ}$ . Find the height of the tower and the width of the canal. [Use $3 = 1.732$ ]

Show answer & solution

Answer: Height

= 103 \approx 17.32

m; width of canal = 10 m

Let height = h and width of canal = $x$ .
From the first point: $tan 6 0^{\circ} = \frac{h}{x}$ , so $h = 3 x$ .
From the second point (20 m farther away): $tan 3 0^{\circ} = \frac{h}{x + 20}$ , so $h = \frac{x + 20}{3}$ .
$3 x = \frac{x + 20}{3} \Rightarrow 3 x = x + 20 \Rightarrow x = 10$ m
$h = 103 = 10 \times 1.732 = 17.32$ m

Q345 marksLong AnswerSome Applications of Trigonometry CBSE 2024 · Basic 430/2/2

From the top of a 7 m high building, the angle of elevation of the top of a cable tower is $6 0^{\circ}$ and the angle of depression of its foot is $4 5^{\circ}$ . Determine the height of the tower. (Use $3 = 1.732$ )

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Answer:

7 (1 + 3) \approx 19.124

Let the building be AB = 7 m and the tower CD; let the horizontal distance be $x$ .
Angle of depression of the foot is $4 5^{\circ}$ : $tan 4 5^{\circ} = \frac{7}{x}$ , so $x = 7$ m.
Angle of elevation of the top is $6 0^{\circ}$ : height of tower above the building's top $= x tan 6 0^{\circ} = 73$ m.
Height of tower $= 7 + 73 = 7 (1 + 1.732) = 19.124$ m

Q325 marksLong AnswerSome Applications of Trigonometry CBSE 2024 · Basic 430/2/3

As observed from the top of a 75 m light house from the sea-level, the angles of depression of two ships are $3 0^{\circ}$ and $4 5^{\circ}$ . If one ship is exactly behind the other on the same side of the light house, find the distance between the two ships. [Use $3 = 1.732$ ]

Show answer & solution

Answer:

75 (3 - 1) \approx 54.9

Let the light house be AB = 75 m. The nearer ship (depression $4 5^{\circ}$ ) is at distance $x$ , the farther one (depression $3 0^{\circ}$ ) at distance $y$ .
$tan 4 5^{\circ} = \frac{75}{x} \Rightarrow x = 75$ m
$tan 3 0^{\circ} = \frac{75}{y} \Rightarrow y = 753$ m
Distance between ships $= 753 - 75 = 75 (1.732 - 1) = 75 \times 0.732 = 54.9$ m

Q345 marksLong AnswerSome Applications of Trigonometry CBSE 2024 · Basic 430/3/1

Two poles of equal height are standing opposite each other on either side of a road, which is 100 m wide. From a point somewhere between them on the road, the angles of elevation of the top of the poles are $6 0^{\circ}$ and $3 0^{\circ}$ respectively. Find the height of the poles and the distances of the point from the poles.

Show answer & solution

Answer: Height

= 253

m (about 43.3 m); distances 25 m and 75 m

Let each pole have height $h$ and the point be $x$ m from the pole with elevation $6 0^{\circ}$ , so $(100 - x)$ m from the other.
$tan 6 0^{\circ} = \frac{h}{x}$ gives $h = 3 x$ .
$tan 3 0^{\circ} = \frac{h}{100 - x}$ gives $h = \frac{100 - x}{3}$ .
$3 x = \frac{100 - x}{3}$ , so $3 x = 100 - x$ and $x = 25$ .
$h = 253 \approx 43.3$ m.
Distances from the poles: 25 m and 75 m.

Q335 marksLong AnswerSome Applications of Trigonometry CBSE 2024 · Basic 430/3/2

A man standing on the deck of a ship, which is 10 m above water level, observes the angle of elevation of the top of a hill as $6 0^{\circ}$ and the angle of depression of the base of the hill as $3 0^{\circ}$ . Find the distance of the hill from the ship and the height of the hill. (Take $3 = 1.73$ )

Show answer & solution

Answer: Distance

= 103 = 17.3

m; height of hill = 40 m

Let the man's eye be at E, 10 m above water, and let the hill be at distance $d$ with height $H$ .
Angle of depression of the base is $3 0^{\circ}$ : $tan 3 0^{\circ} = \frac{10}{d}$ , so $d = 103 = 17.3$ m.
Angle of elevation of the top is $6 0^{\circ}$ : height of top above eye level $= d tan 6 0^{\circ} = 103 \times 3 = 30$ m.
Height of hill $H = 30 + 10 = 40$ m.

Q345 marksLong AnswerSome Applications of Trigonometry CBSE 2024 · Basic 430/3/3

The angle of elevation of a helicopter in air from a point A on the ground is $4 5^{\circ}$ . After a flight of 25 seconds, the angle of elevation changes to $3 0^{\circ}$ . If the helicopter is flying at a constant height of 2500 m, find the speed of the helicopter. (Use $3 = 1.73$ )

Show answer & solution

Answer: 73 m/s (= 262.8 km/h)

Let the first and second positions be above points C and D on the ground, with $A C$ and $A D$ the horizontal distances.
$tan 4 5^{\circ} = \frac{2500}{A C}$ , so $A C = 2500$ m.
$tan 3 0^{\circ} = \frac{2500}{A D}$ , so $A D = 25003$ m.
Distance flown $= A D - A C = 2500 (3 - 1) = 2500 \times 0.73 = 1825$ m.
Speed $= \frac{1825}{25} = 73$ m/s $= 262.8$ km/h.

Q335 marksLong AnswerSome Applications of Trigonometry CBSE 2024 · Basic 430/4/1

A contractor plans to install two slides for the children to play in a park. For the children below the age of 6 years, he prefers to have a slide whose top is at a height of 2.0 m and is inclined at an angle of $3 0^{\circ}$ to the ground, whereas for older children, he wants to have a steep slide at a height of 4.0 m and inclined at an angle of $6 0^{\circ}$ to the ground. What would be the length of the slide in each case ?

Show answer & solution

Answer: 4 m and

\frac{8 3}{3}

m (about 4.62 m)

Slide 1: height 2 m, angle $3 0^{\circ}$ ; length $l_{1}$ satisfies $sin 3 0^{\circ} = \frac{2}{l _{1}}$ .
$\frac{1}{2} = \frac{2}{l _{1}} \Rightarrow l_{1} = 4$ m.
Slide 2: height 4 m, angle $6 0^{\circ}$ ; $sin 6 0^{\circ} = \frac{4}{l _{2}}$ .
$\frac{3}{2} = \frac{4}{l _{2}} \Rightarrow l_{2} = \frac{8}{3} = \frac{8 3}{3} \approx 4.62$ m.

Q325 marksLong AnswerSome Applications of Trigonometry CBSE 2024 · Basic 430/4/2

From a point P on the ground, the angle of elevation of the top of a 15 m tall building is $3 0^{\circ}$ . A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from P is $4 5^{\circ}$ . Find the length of the flagstaff and the distance of the building from the point P. (Use $3 = 1.732$ )

Show answer & solution

Answer: Flagstaff

= 15 (3 - 1) = 10.98

m; distance

= 153 = 25.98

Let the building be AB (B on the ground), AB = 15 m, flagstaff AD $= h$ m, and PB $= x$ m.
In $△ P B A$ : $tan 3 0^{\circ} = \frac{15}{x} \Rightarrow x = 153 = 15 \times 1.732 = 25.98$ m.
In $△ P B D$ : $tan 4 5^{\circ} = \frac{15 + h}{x} \Rightarrow 15 + h = x = 153$ .
$h = 153 - 15 = 15 (3 - 1) = 15 \times 0.732 = 10.98$ m.
Flagstaff is 10.98 m long; the building is 25.98 m from P.

Q355 marksLong AnswerSome Applications of Trigonometry CBSE 2024 · Basic 430/4/3

A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is $6 0^{\circ}$ and from the same point, the angle of elevation of the top of the pedestal is $4 5^{\circ}$ . Find the height of the pedestal. (Use $3 = 1.732$ )

Show answer & solution

Answer:

0.8 (3 + 1) \approx 2.19

Let the pedestal height be $h$ m and the distance of the point from its foot be $x$ m.
$tan 4 5^{\circ} = \frac{h}{x} \Rightarrow x = h$ .
$tan 6 0^{\circ} = \frac{h + 1.6}{x} \Rightarrow 3 h = h + 1.6$ .
$h = \frac{1.6}{3 - 1} = \frac{1.6 ( 3 + 1 )}{2} = 0.8 (3 + 1)$ .
$h = 0.8 \times 2.732 = 2.1856 \approx 2.19$ m.

Q355 marksLong AnswerSome Applications of Trigonometry CBSE 2024 · Standard 30/1/1

A pole 6m high is fixed on the top of a tower. The angle of elevation of the top of the pole observed from a point P on the ground is $6 0^{\circ}$ and the angle of depression of the point P from the top of the tower is $4 5^{\circ}$ . Find the height of the tower and the distance of point P from the foot of the tower. (Use $3 = 1.73$ )

Show answer & solution

Answer: Height of tower = 8.19 m; distance of P from the foot = 8.19 m

Let the tower be BC of height $h$ m with foot B, and pole CD = 6 m; let PB = $d$ m.
Angle of depression of P from C is $4 5^{\circ}$ , so $∠ C P B = 4 5^{\circ}$ and $tan 4 5^{\circ} = \frac{h}{d}$ , giving $d = h$ .
In $△ D B P$ : $tan 6 0^{\circ} = \frac{h + 6}{d}$ , so $3 h = h + 6$ .
$h = \frac{6}{3 - 1} = \frac{6 ( 3 + 1 )}{2} = 3 (3 + 1) = 3 \times 2.73 = 8.19$ m.
Distance PB $= d = h = 8.19$ m.

Q335 marksLong AnswerSome Applications of Trigonometry CBSE 2024 · Standard 30/1/2

From the top of a building 60 m high, the angles of depression of the top and bottom of the vertical lamp post are observed to be $3 0^{\circ}$ and $6 0^{\circ}$ respectively.
(i) Find the horizontal distance between the building and the lamp post.
(ii) Find the distance between the tops of the building and the lamp post.

Show answer & solution

Answer: (i)

203

\approx 34.64

m (ii) 40 m

Let the building be AB = 60 m (top A) and the lamp post CD (top D, foot C), with BC = $d$ m.
(i) Angle of depression of C is $6 0^{\circ}$ : $tan 6 0^{\circ} = \frac{60}{d}$ , so $d = \frac{60}{3} = 203 \approx 34.64$ m.
(ii) Angle of depression of D is $3 0^{\circ}$ : the vertical drop from A to the level of D is $d tan 3 0^{\circ} = 203 \times \frac{1}{3} = 20$ m (so the lamp post is $60 - 20 = 40$ m high).
Distance AD $= \frac{d}{c o s 3 0 ^{\circ}} = \frac{20 3}{3 /2} = 40$ m (check: $(203)^{2} + 2 0^{2} = 1600 = 40$ ).

Q325 marksLong AnswerSome Applications of Trigonometry CBSE 2024 · Standard 30/1/3

From the top of a 15 m high building, the angle of elevation of the top of a tower is found to be $3 0^{\circ}$ . From the bottom of the same building, the angle of elevation of the top of the tower is found to be $6 0^{\circ}$ . Find the height of the tower and the distance between tower and the building.

Show answer & solution

Answer: Height of tower = 22.5 m; distance

= \frac{15 3}{2} \approx 12.99

Let the tower have height $H$ m and be $d$ m from the building.
From the bottom of the building: $tan 6 0^{\circ} = \frac{H}{d}$ , so $H = 3 d$ ... (1)
From the top (15 m): $tan 3 0^{\circ} = \frac{H - 15}{d}$ , so $H - 15 = \frac{d}{3}$ ... (2)
Substituting (1) in (2): $3 d - \frac{d}{3} = 15$ , i.e. $\frac{2 d}{3} = 15$ , so $d = \frac{15 3}{2} \approx 12.99$ m.
$H = 3 \times \frac{15 3}{2} = 22.5$ m.

Q345 marksLong AnswerSome Applications of Trigonometry CBSE 2024 · Standard 30/2/1

From the top of a 45 m high light house, the angles of depression of two ships, on the opposite side of it, are observed to be $3 0^{\circ}$ and $6 0^{\circ}$ . If the line joining the ships passes through the foot of the light house, find the distance between the ships. (Use $3 = 1.73$ )

Show answer & solution

Answer:

603

m = 103.8 m

Let AB = 45 m be the light house (B the foot) and C, D the ships on opposite sides of B, with depression angles $3 0^{\circ}$ and $6 0^{\circ}$ .
Angle of depression = angle of elevation from the ship (alternate angles).
In $△ A B C$ : $tan 3 0^{\circ} = \frac{45}{B C}$ , so $B C = 453$ m.
In $△ A B D$ : $tan 6 0^{\circ} = \frac{45}{B D}$ , so $B D = \frac{45}{3} = 153$ m.
Distance $C D = B C + B D = 603 = 60 \times 1.73 = 103.8$ m

Q325 marksLong AnswerSome Applications of Trigonometry CBSE 2024 · Standard 30/2/2

The angle of elevation of an aircraft from a point A on the ground is $6 0^{\circ}$ . After a flight of 30 seconds, the angle of elevation changes to $3 0^{\circ}$ . The aircraft is flying at a constant height of $35003$ m at a uniform speed. Find the speed of the aircraft.

Show answer & solution

Answer:

\frac{700}{3}

m/s (about 233.33 m/s, i.e. 840 km/h)

Let B and C be the first and second positions of the aircraft, and M, N the points on the ground vertically below them, so BM = CN = $35003$ m.
In right $△ A M B$ : $tan 6 0^{\circ} = \frac{B M}{A M}$ , so $A M = \frac{3500 3}{3} = 3500$ m.
In right $△ A N C$ : $tan 3 0^{\circ} = \frac{C N}{A N}$ , so $A N = 35003 \times 3 = 10500$ m.
Distance flown $B C = M N = 10500 - 3500 = 7000$ m in 30 s.
Speed $= \frac{7000}{30} = \frac{700}{3}$ m/s $\approx 233.33$ m/s ( $= 840$ km/h).

Q325 marksLong AnswerSome Applications of Trigonometry CBSE 2024 · Standard 30/2/3

A person standing on the bank of a river observes that the angle of elevation of the top of a tower on the opposite bank is $6 0^{\circ}$ . When he moves 30 m away from the bank, he finds the angle of elevation to be $3 0^{\circ}$ . Find the height of the tower and width of the river. (Take $3 = 1.732$ )

Show answer & solution

Answer: Height of tower

= 153

m = 25.98 m; width of river = 15 m

Let AB = $h$ m be the tower (B its foot), C the first position on the bank and D the point 30 m further away, with BC = $x$ m.
In right $△ A B C$ : $tan 6 0^{\circ} = \frac{h}{x}$ , so $h = 3 x$ ... (1)
In right $△ A B D$ : $tan 3 0^{\circ} = \frac{h}{x + 30}$ , so $h = \frac{x + 30}{3}$ ... (2)
From (1) and (2): $3 x = x + 30$ , so $x = 15$ m.
$h = 153 = 15 \times 1.732 = 25.98$ m

Q345 marksLong AnswerSome Applications of Trigonometry CBSE 2024 · Standard 30/3/1

From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are $4 5^{\circ}$ and $6 0^{\circ}$ respectively. Find the height of the tower.

Show answer & solution

Answer:

20 (3 - 1)

\approx 14.64

Let the point be P, the foot of the building B, its top C (BC = 20 m), and the top of the tower D (CD = $h$ ).
In right $△ P B C$ : $tan 4 5^{\circ} = \frac{20}{P B} \Rightarrow P B = 20$ m.
In right $△ P B D$ : $tan 6 0^{\circ} = \frac{20 + h}{20} \Rightarrow 20 + h = 203$ .
$h = 203 - 20 = 20 (3 - 1) \approx 20 \times 0.732 = 14.64$ m.

Q355 marksLong AnswerSome Applications of Trigonometry CBSE 2024 · Standard 30/3/2

The angle of elevation of a jet plane from a point A on the ground is $6 0^{\circ}$ . After a flight of 30 seconds, the angle of elevation changes to $3 0^{\circ}$ . If the jet plane is flying at a constant height of $36003$ m, find the speed of the jet plane.

Show answer & solution

Answer: 240 m/s (= 864 km/h)

Let the plane be at P then Q, with feet of perpendiculars B and C on the ground line through A; PB = QC = $36003$ m.
$tan 6 0^{\circ} = \frac{3600 3}{A B} \Rightarrow A B = 3600$ m.
$tan 3 0^{\circ} = \frac{3600 3}{A C} \Rightarrow A C = 36003 \times 3 = 10800$ m.
Distance flown PQ = BC = 10800 - 3600 = 7200 m.
Speed = $\frac{7200}{30} = 240$ m/s = $240 \times \frac{18}{5} = 864$ km/h.

Q325 marksLong AnswerSome Applications of Trigonometry CBSE 2024 · Standard 30/3/3

A man on a cliff observes a boat at an angle of depression of $3 0^{\circ}$ which is approaching the shore to the point immediately beneath the observer with a uniform speed. Six minutes later, the angle of depression of the boat is found to be $6 0^{\circ}$ . Find the time taken by the boat from here to reach the shore.

Show answer & solution

Answer: 3 minutes

Let the height of the cliff be $h$ and B the point on the shore beneath the man.
First position C: $tan 3 0^{\circ} = \frac{h}{B C} \Rightarrow B C = h 3$ .
Second position D: $tan 6 0^{\circ} = \frac{h}{B D} \Rightarrow B D = \frac{h}{3}$ .
Distance covered in 6 minutes: $C D = h 3 - \frac{h}{3} = \frac{2 h}{3}$ .
Speed = $\frac{2 h}{6 3} = \frac{h}{3 3}$ per minute.
Time for the remaining distance BD = $\frac{h}{3} \div \frac{h}{3 3} = 3$ minutes.

Q325 marksLong AnswerSome Applications of Trigonometry CBSE 2024 · Standard 30/4/1

Two pillars of equal lengths stand on either side of a road which is 100 m wide, exactly opposite to each other. At a point on the road between the pillars, the angles of elevation of the tops of the pillars are $6 0^{\circ}$ and $3 0^{\circ}$ . Find the length of each pillar and distance of the point on the road from the pillars. (Use $3$ = 1.732)

Show answer & solution

Answer: Each pillar is

253

m = 43.3 m long; the point is 25 m from the pillar seen at

6 0^{\circ}

and 75 m from the other pillar.

Let each pillar have height $h$ and let the point be $x$ m from the pillar seen at $6 0^{\circ}$ , so $100 - x$ m from the other.
$tan 6 0^{\circ} = \frac{h}{x}$ gives $h = 3 x$ .
$tan 3 0^{\circ} = \frac{h}{100 - x}$ gives $h = \frac{100 - x}{3}$ .
$3 x = \frac{100 - x}{3}$ , so $3 x = 100 - x$ and $x = 25$ m.
$h = 253 = 25 \times 1.732 = 43.3$ m.
Distances from the pillars: 25 m and 75 m.

Q355 marksLong AnswerSome Applications of Trigonometry CBSE 2024 · Standard 30/4/2

The angles of depression of the top and the bottom of a 50 m high building from the top of a tower are $4 5^{\circ}$ and $6 0^{\circ}$ , respectively. Find the height of the tower. (Use $3$ = 1.73)

Show answer & solution

Answer:

25 (3 + 3)

m = 118.25 m

Let the tower have height $h$ m and let the horizontal distance between the tower and the building be $x$ m.
Bottom of building ( $6 0^{\circ}$ ): $tan 6 0^{\circ} = \frac{h}{x}$ , so $h = 3 x$ .
Top of building ( $4 5^{\circ}$ ): $tan 4 5^{\circ} = \frac{h - 50}{x}$ , so $x = h - 50$ .
$h = 3 (h - 50)$ , so $h (3 - 1) = 503$ .
$h = \frac{50 3}{3 - 1} = \frac{50 3 ( 3 + 1 )}{2} = 25 (3 + 3)$ .
$h = 25 \times 4.73 = 118.25$ m.

Q355 marksLong AnswerSome Applications of Trigonometry CBSE 2024 · Standard 30/4/3

The angles of depression of the top and the bottom of a 8 m tall building from the top of a multi-storeyed building are $3 0^{\circ}$ and $4 5^{\circ}$ respectively. Find the height of the multi-storeyed building and the distance between the two buildings.

Show answer & solution

Answer: Height = distance =

4 (3 + 3)

\approx

18.93 m

Let the multi-storeyed building have height $h$ m and let the distance between the buildings be $x$ m.
Bottom of the 8 m building ( $4 5^{\circ}$ ): $tan 4 5^{\circ} = \frac{h}{x}$ , so $x = h$ .
Top of the 8 m building ( $3 0^{\circ}$ ): $tan 3 0^{\circ} = \frac{h - 8}{x}$ , so $h - 8 = \frac{h}{3}$ .
$h (3 - 1) = 83$ , so $h = \frac{8 3}{3 - 1} = \frac{8 3 ( 3 + 1 )}{2} = 4 (3 + 3)$ m.
$h = 12 + 43 \approx 18.93$ m, and the distance $x = h = 4 (3 + 3) \approx 18.93$ m.

Q335 marksLong AnswerSome Applications of Trigonometry CBSE 2024 · Standard 30/5/1

From a point on a bridge across the river, the angles of depressions of the banks on opposite sides of the river are $3 0^{\circ}$ and $6 0^{\circ}$ respectively. If the bridge is at a height of 4 m from the banks, find the width of the river.

Show answer & solution

Answer:

\frac{16 3}{3}

m (about 9.24 m)

Let P be the point on the bridge, PD = 4 m its height above the line of the banks, and A, B the banks on opposite sides of D.
Angles of depression equal the angles of elevation: $∠ P A D = 3 0^{\circ}$ , $∠ P B D = 6 0^{\circ}$ .
$tan 3 0^{\circ} = \frac{4}{A D} \Rightarrow A D = 43$ m.
$tan 6 0^{\circ} = \frac{4}{B D} \Rightarrow B D = \frac{4}{3} = \frac{4 3}{3}$ m.
Width $A B = 43 + \frac{4 3}{3} = \frac{16 3}{3}$ m $\approx 9.24$ m.

Also asked in: 2024 Standard 30/5/2

Q345 marksLong AnswerSome Applications of Trigonometry CBSE 2024 · Standard 30/5/3

From a window 15 metres high above the ground in a street, the angles of elevation and depression of the top and the foot of another house on the opposite side of the street are $3 0^{\circ}$ and $4 5^{\circ}$ respectively. Find the height of the opposite house. (Use $3$ = 1.732)

Show answer & solution

Answer:

15 + 53 \approx 23.66

Let W be the window, 15 m above the ground, and let the opposite house be CD with foot D and top C. Draw WE $⊥$ CD.
Angle of depression of D is $4 5^{\circ}$ : $tan 4 5^{\circ} = \frac{15}{width}$ , so the width of the street WE = 15 m, and ED = 15 m.
Angle of elevation of C is $3 0^{\circ}$ : $tan 3 0^{\circ} = \frac{C E}{15}$ , so $C E = \frac{15}{3} = 53 = 8.66$ m.
Height of the house $= C E + E D = 8.66 + 15 = 23.66$ m.

Q355 marksLong AnswerSome Applications of Trigonometry CBSE 2023 · Basic 430/1/1

A TV tower stands vertically on the bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is $6 0^{\circ}$ . From another point 20 m away from the point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is $3 0^{\circ}$ . Find the height of the tower.

Show answer & solution

Answer:

103

m ≈ 17.32 m

Let the tower AB have height $h$ m, and let the width of the canal be $x$ m (point C opposite the tower).
In $△ A B C$ : $tan 6 0^{\circ} = \frac{h}{x}$ , so $h = 3 x$ .
Point D is 20 m beyond C. In $△ A B D$ : $tan 3 0^{\circ} = \frac{h}{x + 20}$ , so $x + 20 = 3 h = 3 x$ .
$2 x = 20$ , so $x = 10$ m.
$h = 103 \approx 17.32$ m

Also asked in: 2023 Basic 430/1/2, 2023 Basic 430/1/3

Q35 (OR) (OR)5 marksLong AnswerSome Applications of Trigonometry CBSE 2023 · Basic 430/1/1

An aeroplane when flying at a height of 4000 m from the ground passes vertically above another aeroplane at an instant when the angles of elevation of the two planes from the same point on the ground are $6 0^{\circ}$ and $4 5^{\circ}$ respectively. Find the vertical distance between the aeroplanes at that instant. (Use $3 = 1.73$ )

Show answer & solution

Answer:

\frac{4000 ( 3 - 3 )}{3}

m ≈ 1693.33 m

Let O be the point on the ground and C the point directly below both planes, with OC = $x$ m.
Upper plane A (height 4000 m): $tan 6 0^{\circ} = \frac{4000}{x}$ , so $x = \frac{4000}{3}$ .
Lower plane B: $tan 4 5^{\circ} = \frac{B C}{x}$ , so $B C = x = \frac{4000}{3} = \frac{4000 3}{3}$ m.
Vertical distance $A B = 4000 - \frac{4000 3}{3} = \frac{4000 ( 3 - 3 )}{3}$
$= \frac{4000 \times 1.27}{3} \approx 1693.33$ m

Also asked in: 2023 Basic 430/1/2, 2023 Basic 430/1/3

Q325 marksLong AnswerSome Applications of Trigonometry CBSE 2023 · Basic 430/2/1

From the top of a building 60 m high, the angles of depression of the top and bottom of a tower are observed to be $3 0^{\circ}$ and $6 0^{\circ}$ respectively. Find the height of the tower. Also, find the distance between the building and the tower. (Use $3 = 1.732$ )

Show answer & solution

Answer: Height of tower = 40 m; distance =

203

m = 34.64 m

Let AB be the building (AB = 60 m), CD the tower of height h and BD = x the distance between them.
Angle of depression of the bottom D is $6 0^{\circ}$ : $tan 6 0^{\circ} = \frac{60}{x}$ , so $x = \frac{60}{3} = 203$ m
Angle of depression of the top C is $3 0^{\circ}$ : $tan 3 0^{\circ} = \frac{60 - h}{x}$
$60 - h = \frac{20 3}{3} = 20$ , so h = 40 m
Distance $= 203 = 20 \times 1.732 = 34.64$ m

Also asked in: 2023 Basic 430/2/2, 2023 Basic 430/2/3

Q32 (OR) (OR)5 marksLong AnswerSome Applications of Trigonometry CBSE 2023 · Basic 430/2/1

The angle of elevation of the top of a building from a point A on the ground is $3 0^{\circ}$ . On moving a distance of 30 m towards its base to the point B, the angle of elevation changes to $4 5^{\circ}$ . Find the height of the building and the distance of its base from point A. (Use $3 = 1.732$ )

Show answer & solution

Answer: Height =

15 (3 + 1)

m = 40.98 m; distance from A = 70.98 m

Let PQ = h be the building with foot Q, and BQ = y.
From B: $tan 4 5^{\circ} = \frac{h}{y}$ , so y = h
From A: $tan 3 0^{\circ} = \frac{h}{y + 30} = \frac{h}{h + 30}$
$3 h = h + 30$ , so $h = \frac{30}{3 - 1} = \frac{30 ( 3 + 1 )}{2} = 15 (3 + 1)$
$h = 15 \times 2.732 = 40.98$ m
Distance of base from A = h + 30 = 70.98 m

Also asked in: 2023 Basic 430/2/2, 2023 Basic 430/2/3

Q325 marksLong AnswerSome Applications of Trigonometry CBSE 2023 · Basic 430/4/1

The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is $3 0^{\circ}$ than when it was $6 0^{\circ}$ . Find the height of the tower.

Show answer & solution

Answer:

203

m (about 34.64 m)

Let the height of the tower be $h$ m.
Shadow at $6 0^{\circ}$ : $\frac{h}{t a n 6 0 ^{\circ}} = \frac{h}{3}$ .
Shadow at $3 0^{\circ}$ : $\frac{h}{t a n 3 0 ^{\circ}} = h 3$ .
$h 3 - \frac{h}{3} = 40 \Rightarrow \frac{3 h - h}{3} = 40$ .
$2 h = 403 \Rightarrow h = 203$ m $\approx 34.64$ m.

Also asked in: 2023 Basic 430/4/2

Q32 (OR) (OR)5 marksLong AnswerSome Applications of Trigonometry CBSE 2023 · Basic 430/4/1

From the top of a 7 m high building, the angle of elevation of the top of a cable tower is $6 0^{\circ}$ and the angle of depression of its foot is $4 5^{\circ}$ . Determine the height of the tower.

Show answer & solution

Answer:

7 (1 + 3)

m (about 19.12 m)

Let AB be the building (AB = 7 m) and CD the tower, with C its foot. Let the horizontal through A meet CD at E, so CE = 7 m.
Angle of depression of C is $4 5^{\circ}$ : $tan 4 5^{\circ} = \frac{7}{B C} \Rightarrow B C = 7$ m, so AE = 7 m.
Angle of elevation of D is $6 0^{\circ}$ : $tan 6 0^{\circ} = \frac{D E}{A E} \Rightarrow D E = 73$ m.
Height of tower $= C E + D E = 7 + 73 = 7 (1 + 3)$ m $\approx 19.12$ m.

Also asked in: 2023 Basic 430/4/2

Q355 marksLong AnswerSome Applications of Trigonometry CBSE 2023 · Basic 430/4/3

As observed from the top of a 75 m high light house from the sea-level, the angles of depression of two ships are $3 0^{\circ}$ and $4 5^{\circ}$ . If one ship is exactly behind the other on the same side of the light house, find the distance between the two ships. (use $3 = 1.73$ )

Show answer & solution

Answer:

75 (3 - 1) = 54.75

Let AB be the light house (AB = 75 m), with ships at C (depression $4 5^{\circ}$ ) and D (depression $3 0^{\circ}$ ) on the same side.
$tan 4 5^{\circ} = \frac{75}{B C} \Rightarrow B C = 75$ m.
$tan 3 0^{\circ} = \frac{75}{B D} \Rightarrow B D = 753$ m.
Distance $C D = B D - B C = 75 (3 - 1) = 75 \times 0.73 = 54.75$ m.

Q35 (OR) (OR)5 marksLong AnswerSome Applications of Trigonometry CBSE 2023 · Basic 430/4/3

From a point P on the ground, the angle of elevation of the top of a 10 m tall building is $3 0^{\circ}$ . A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from P is $4 5^{\circ}$ . Find the length of the flagstaff and the distance of the building from the point P. (use $3 = 1.73$ )

Show answer & solution

Answer: Flagstaff

= 10 (3 - 1) = 7.3

m; distance

= 103 = 17.3

Let the building be AB (AB = 10 m), the flagstaff BD on top, and P on the ground with PA horizontal.
$tan 3 0^{\circ} = \frac{10}{P A} \Rightarrow P A = 103 = 17.3$ m.
$tan 4 5^{\circ} = \frac{A D}{P A} \Rightarrow A D = P A = 103$ m.
Flagstaff $B D = A D - A B = 103 - 10 = 10 (0.73) = 7.3$ m.

Q325 marksLong AnswerSome Applications of Trigonometry CBSE 2023 · Basic 430/5/1

A person walking 48 m towards a tower in a horizontal line through its base observes that angle of elevation of the top of the tower changes from $4 5^{\circ}$ to $6 0^{\circ}$ . Find the height of the tower and distance of the person, now, from the tower. (Use $3$ = 1.732)

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Answer: Height = 113.568 m (about 113.57 m); present distance = 65.568 m (about 65.57 m)

Let the height be $h$ and the present distance be $x$ m; the earlier distance was $(x + 48)$ m.
$tan 6 0^{\circ} = \frac{h}{x}$ , so $h = 3 x$ .
$tan 4 5^{\circ} = \frac{h}{x + 48}$ , so $h = x + 48$ .
$3 x = x + 48$ , so $x = \frac{48}{3 - 1} = 24 (3 + 1) = 24 \times 2.732 = 65.568$ m.
$h = x + 48 = 113.568$ m.
Height of the tower ≈ 113.57 m; the person is now ≈ 65.57 m from the tower.

Q345 marksLong AnswerSome Applications of Trigonometry CBSE 2023 · Basic 430/5/2

A man in a boat rowing away from a lighthouse 100 m high takes 2 minutes to change the angle of elevation of the top of lighthouse from $6 0^{\circ}$ to $4 5^{\circ}$ . Find the speed of the boat. (Use $3$ = 1.73)

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Answer:

\frac{127}{6} \approx 21.17

m/min (about 1.27 km/h)

Let the lighthouse be AB = 100 m, with the boat first at C ( $6 0^{\circ}$ ) and then at D ( $4 5^{\circ}$ ).
$tan 6 0^{\circ} = \frac{100}{B C}$ , so BC $= \frac{100}{3} = \frac{100 3}{3} = \frac{173}{3}$ m.
$tan 4 5^{\circ} = \frac{100}{B D}$ , so BD = 100 m.
Distance rowed CD $= 100 - \frac{173}{3} = \frac{127}{3} \approx 42.33$ m.
Speed $= \frac{127}{3} \div 2 = \frac{127}{6} \approx 21.17$ m/min ( $= 1270$ m/h $= 1.27$ km/h).

Q345 marksLong AnswerSome Applications of Trigonometry CBSE 2023 · Basic 430/5/3

From the top of a building 50 m high, the angles of depression of the top and bottom of a tower are observed to be $3 0^{\circ}$ and $6 0^{\circ}$ . Find the height of the tower and distance between the building and the tower. (Take $3$ = 1.73)

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Answer: Height of tower

= \frac{100}{3} \approx 33.33

m; distance

= \frac{50 3}{3} \approx 28.83

Let the building be AB = 50 m and the tower CD = $h$ m, at distance BD = $x$ m.
Bottom of tower (depression $6 0^{\circ}$ ): $tan 6 0^{\circ} = \frac{50}{x}$ , so $x = \frac{50}{3} = \frac{50 3}{3} = \frac{50 \times 1.73}{3} \approx 28.83$ m.
Top of tower (depression $3 0^{\circ}$ ): $tan 3 0^{\circ} = \frac{50 - h}{x}$ , so $50 - h = \frac{x}{3} = \frac{50}{3}$ .
$h = 50 - \frac{50}{3} = \frac{100}{3} \approx 33.33$ m.
Height of the tower ≈ 33.33 m; distance between the building and the tower ≈ 28.83 m.

Q325 marksLong AnswerSome Applications of Trigonometry CBSE 2023 · Basic 430/6/1

From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are $3 0^{\circ}$ and $4 5^{\circ}$ respectively. If the bridge is at a height of 3 m from the banks, find the width of the river. (Use $3 = 1.73$ )

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Answer: 8.19 m

Let P be the point on the bridge, 3 m above the foot D on the line joining the banks A and B, with A and B on opposite sides of D.
Angle of depression of A is $3 0^{\circ}$ : $tan 3 0^{\circ} = \frac{3}{A D}$ , so $A D = 33$ m.
Angle of depression of B is $4 5^{\circ}$ : $tan 4 5^{\circ} = \frac{3}{D B}$ , so $D B = 3$ m.
Width $A B = 33 + 3 = 3 (1.73 + 1) = 8.19$ m.

Also asked in: 2023 Basic 430/6/2, 2023 Basic 430/6/3

Q32 (OR) (OR)5 marksLong AnswerSome Applications of Trigonometry CBSE 2023 · Basic 430/6/1

From a point on the ground, the angle of elevation of the bottom and top of a transmission tower fixed at the top of a 20 m high building are $4 5^{\circ}$ and $6 0^{\circ}$ respectively. Find the height of the tower. (Use $3 = 1.73$ )

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Answer: 14.6 m

Let P be the point on the ground at distance $x$ m from the foot of the building, building height 20 m and tower height $h$ m.
Bottom of tower: $tan 4 5^{\circ} = \frac{20}{x}$ , so $x = 20$ m.
Top of tower: $tan 6 0^{\circ} = \frac{20 + h}{20}$ , so $20 + h = 203$ .
$h = 20 (3 - 1) = 20 \times 0.73 = 14.6$ m.

Also asked in: 2023 Basic 430/6/2, 2023 Basic 430/6/3

Q335 marksLong AnswerSome Applications of Trigonometry CBSE 2023 · Standard 30/1/1

The angle of elevation of the top of a tower 24 m high from the foot of another tower in the same plane is $6 0^{\circ}$ . The angle of elevation of the top of second tower from the foot of the first tower is $3 0^{\circ}$ . Find the distance between two towers and the height of the other tower. Also, find the length of the wire attached to the tops of both the towers.

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Answer: Distance

= 83

m, height of other tower = 8 m, wire

= 87

Let AB = 24 m be the first tower, CD = h the second, and BD = d the distance between their feet.
From D: $tan 6 0^{\circ} = \frac{24}{d}$ , so $d = \frac{24}{3} = 83$ m.
From B: $tan 3 0^{\circ} = \frac{h}{d}$ , so $h = \frac{8 3}{3} = 8$ m.
Wire AC: $A C^{2} = d^{2} + (24 - 8)^{2} = 192 + 256 = 448$ .
$A C = 448 = 87$ m.

Q33 (OR) (OR)5 marksLong AnswerSome Applications of Trigonometry CBSE 2023 · Standard 30/1/1

A spherical balloon of radius r subtends an angle of $6 0^{\circ}$ at the eye of an observer. If the angle of elevation of its centre is $4 5^{\circ}$ from the same point, then prove that height of the centre of the balloon is $2$ times its radius.

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Answer: Proved.

Let E be the eye, C the centre of the balloon, and EA a tangent from E touching the balloon at A.
The two tangents from E make $6 0^{\circ}$ , so $∠ A E C = 3 0^{\circ}$ ; also CA $⊥$ EA, CA = r.
$sin 3 0^{\circ} = \frac{r}{E C}$ , so EC = 2r.
Let CD be the perpendicular from C to the horizontal through E. $∠ C E D = 4 5^{\circ}$ .
$C D = E C sin 4 5^{\circ} = 2 r \times \frac{1}{2} = 2 r$ .
Hence the height of the centre is $2$ times the radius.

Q325 marksLong AnswerSome Applications of Trigonometry CBSE 2023 · Standard 30/1/2

A ladder set against a wall at an angle $4 5^{\circ}$ to the ground. If the foot of the ladder is pulled away from the wall through a distance of 4 m, its top slides a distance of 3 m down the wall making an angle $3 0^{\circ}$ with the ground. Find the final height of the top of the ladder from the ground and length of the ladder.

Show answer & solution

Answer: Final height

= \frac{7 ( 3 + 1 )}{2}

m ≈ 9.56 m; length of ladder

= 7 (3 + 1)

m ≈ 19.12 m

Let the initial height of the top be h m. At $4 5^{\circ}$ the foot is also h m from the wall.
Finally the top is at (h – 3) m and the foot at (h + 4) m from the wall, with angle $3 0^{\circ}$ .
$tan 3 0^{\circ} = \frac{h - 3}{h + 4}$ , so $3 (h - 3) = h + 4$ , $h = \frac{4 + 3 3}{3 - 1} = \frac{13 + 7 3}{2}$ .
Final height $= h - 3 = \frac{7 + 7 3}{2} = \frac{7 ( 3 + 1 )}{2}$ m ≈ 9.56 m.
Length of ladder $= \frac{h - 3}{s i n 3 0 ^{\circ}} = 7 (3 + 1)$ m ≈ 19.12 m.

Q355 marksLong AnswerSome Applications of Trigonometry CBSE 2023 · Standard 30/1/3

An aeroplane when flying at a height of 3000 m from the ground passes vertically above another aeroplane at an instant when the angles of elevation of the two planes from the same point on the ground are $6 0^{\circ}$ and $4 5^{\circ}$ respectively. Find the vertical distance between the aeroplanes at that instant. Also, find the distance of first plane from the point of observation. (Take $3 = 1.73$ )

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Answer: Vertical distance

= 1000 (3 - 3)

m = 1270 m; distance of first plane

= 20003

m = 3460 m

Let O be the observation point, C the point on the ground below the planes, A the first plane (AC = 3000 m) and B the second plane.
$tan 6 0^{\circ} = \frac{3000}{O C}$ , so $O C = \frac{3000}{3} = 10003$ m.
$tan 4 5^{\circ} = \frac{B C}{O C}$ , so $B C = 10003 = 1730$ m.
Vertical distance AB = 3000 – 1730 = 1270 m.
$O A = \frac{A C}{s i n 6 0 ^{\circ}} = \frac{3000 \times 2}{3} = 20003 = 3460$ m.

Q335 marksLong AnswerSome Applications of Trigonometry CBSE 2023 · Standard 30/2/1

As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are $3 0^{\circ}$ and $6 0^{\circ}$ . If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
(Use $3 = 1.73$ )

Show answer & solution

Answer:

503

m = 86.5 m

Let the lighthouse be AB = 75 m, with ships at C (depression $6 0^{\circ}$ ) and D (depression $3 0^{\circ}$ ).
$tan 6 0^{\circ} = \frac{75}{B C} \Rightarrow B C = \frac{75}{3} = 253$ m
$tan 3 0^{\circ} = \frac{75}{B D} \Rightarrow B D = 753$ m
CD $= 753 - 253 = 503 = 50 \times 1.73 = 86.5$ m

Also asked in: 2023 Standard 30/2/3

Q33 (OR) (OR)5 marksLong AnswerSome Applications of Trigonometry CBSE 2023 · Standard 30/2/1

From a point on the ground, the angle of elevation of the bottom and top of a transmission tower fixed at the top of 30 m high building are $3 0^{\circ}$ and $6 0^{\circ}$ , respectively. Find the height of the transmission tower. (Use $3 = 1.73$ )

Show answer & solution

Answer: 60 m

Let the point be P, distance $x$ from the foot of the building, and tower height $h$ .
$tan 3 0^{\circ} = \frac{30}{x} \Rightarrow x = 303$ m
$tan 6 0^{\circ} = \frac{30 + h}{x} \Rightarrow 30 + h = 3 \times 303 = 90$
$h = 60$ m

Also asked in: 2023 Standard 30/2/3

Q355 marksLong AnswerSome Applications of Trigonometry CBSE 2023 · Standard 30/2/2

The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is $6 0^{\circ}$ and the angle of elevation of the top of the second tower from the foot of the first tower is $3 0^{\circ}$ . Find the distance between the two towers and also the height of the other tower.

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Answer: Distance

= 103

m (about 17.32 m); height of the other tower = 10 m

Let AB = 30 m be the first tower and CD = $h$ the other tower, with BD = $x$ the distance between their feet.
From D: $tan 6 0^{\circ} = \frac{30}{x} \Rightarrow x = \frac{30}{3} = 103$ m
From B: $tan 3 0^{\circ} = \frac{h}{x} \Rightarrow h = 103 \times \frac{1}{3} = 10$ m

Q35 (OR) (OR)5 marksLong AnswerSome Applications of Trigonometry CBSE 2023 · Standard 30/2/2

From the top of a tower 100 m high, a man observes two cars on the opposite sides of the tower with angles of depression $3 0^{\circ}$ and $4 5^{\circ}$ respectively. Find the distance between the two cars. (Use $3 = 1.73$ )

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Answer:

100 (3 + 1)

m = 273 m

Let the tower be AB = 100 m and the cars be C and D on opposite sides.
Depression $3 0^{\circ}$ : $tan 3 0^{\circ} = \frac{100}{B C} \Rightarrow B C = 1003$ m
Depression $4 5^{\circ}$ : $tan 4 5^{\circ} = \frac{100}{B D} \Rightarrow B D = 100$ m
CD $= 1003 + 100 = 100 (1.73 + 1) = 273$ m

Q325 marksLong AnswerSome Applications of Trigonometry CBSE 2023 · Standard 30/4/1

A straight highway leads to the foot of a tower. A man standing on the top of the 75 m high tower observes two cars at angles of depression of $3 0^{\circ}$ and $6 0^{\circ}$ , which are approaching the foot of the tower. If one car is exactly behind the other on the same side of the tower, find the distance between the two cars. (use $3 = 1.73$ )

Show answer & solution

Answer:

503

m = 86.5 m

Let AB = 75 m be the tower, and C, D the cars with angles of depression $6 0^{\circ}$ and $3 0^{\circ}$ (C nearer).
Angle of elevation from each car equals the angle of depression.
In $△ A B C$ : $tan 6 0^{\circ} = \frac{75}{B C} \Rightarrow B C = \frac{75}{3} = 253$ m
In $△ A B D$ : $tan 3 0^{\circ} = \frac{75}{B D} \Rightarrow B D = 753$ m
CD $= B D - B C = 503 = 50 \times 1.73 = 86.5$ m

Also asked in: 2023 Standard 30/4/2, 2023 Standard 30/4/3

Q32 (OR) (OR)5 marksLong AnswerSome Applications of Trigonometry CBSE 2023 · Standard 30/4/1

From the top of a 7 m high building, the angle of elevation of the top of a cable tower is $6 0^{\circ}$ and the angle of depression of its foot is $3 0^{\circ}$ . Determine the height of the tower.

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Answer: 28 m

Let AB = 7 m be the building and CD the tower; draw AE $⊥$ CD, so ED = AB = 7 m and AE = BD.
Angle of depression of D is $3 0^{\circ}$ : $tan 3 0^{\circ} = \frac{A B}{B D} \Rightarrow B D = 73$ m
Angle of elevation of C is $6 0^{\circ}$ : $tan 6 0^{\circ} = \frac{C E}{A E} \Rightarrow C E = 73 \times 3 = 21$ m
Height of tower $= C E + E D = 21 + 7 = 28$ m

Also asked in: 2023 Standard 30/4/2, 2023 Standard 30/4/3

Q325 marksLong AnswerSome Applications of Trigonometry CBSE 2023 · Standard 30/5/1

One observer estimates the angle of elevation to the basket of a hot air balloon to be $6 0^{\circ}$ , while another observer 100 m away estimates the angle of elevation to be $3 0^{\circ}$ . Find :
(a) The height of the basket from the ground.
(b) The distance of the basket from the first observer's eye.
(c) The horizontal distance of the second observer from the basket.

Show answer & solution

Answer: (a)

503

m ≈ 86.6 m (b) 100 m (c) 150 m

Let the basket be at height h above the point M on the ground. First observer at A, second at B, both on the same side, AB = 100 m, AM = x.
$tan 6 0^{\circ} = \frac{h}{x}$ , so $h = 3 x$ .
$tan 3 0^{\circ} = \frac{h}{x + 100}$ , so $x + 100 = 3 h = 3 x$ , giving x = 50 m.
(a) $h = 503 \approx 86.6$ m.
(b) Distance from first observer $= \frac{x}{c o s 6 0 ^{\circ}} = \frac{50}{1/2} = 100$ m.
(c) Horizontal distance of second observer = x + 100 = 150 m.

Q345 marksLong AnswerSome Applications of Trigonometry CBSE 2023 · Standard 30/5/2

The angle of elevation of the top of a vertical tower from a point P on the ground is $6 0^{\circ}$ . From another point Q, 10 m vertically above the first point P, its angle of elevation is $3 0^{\circ}$ . Find :
(a) The height of the tower.
(b) The distance of the point P from the foot of the tower.
(c) The distance of the point P from the top of the tower.

Show answer & solution

Answer: (a) 15 m (b)

53

m ≈ 8.66 m (c)

103

m ≈ 17.32 m

Let the tower AB have height h (B the foot) and PB = x.
From P: $tan 6 0^{\circ} = \frac{h}{x}$ , so $h = 3 x$ .
From Q (10 m above P): $tan 3 0^{\circ} = \frac{h - 10}{x}$ , so $h - 10 = \frac{x}{3}$ .
$3 x - 10 = \frac{x}{3}$ gives $3 x - x = 103$ , so $x = 53$ m.
(a) $h = 3 \times 53 = 15$ m.
(b) PB $= 53 \approx 8.66$ m.
(c) PA $= \frac{x}{c o s 6 0 ^{\circ}} = 103 \approx 17.32$ m.

Q355 marksLong AnswerSome Applications of Trigonometry CBSE 2023 · Standard 30/5/3

Two pillars are standing on either side of a 80 m wide road. Height of one pillar is 20 m more than the height of the other pillar. From a point on the road between the pillars, the angle of elevation of the higher pillar is $6 0^{\circ}$ , whereas that of the other pillar is $3 0^{\circ}$ . Find the position of the point between the pillars and the height of each pillar. (Use $3 = 1.73$ )

Show answer & solution

Answer: The point is

20 + 53 \approx 28.65

m from the higher pillar (51.35 m from the other); heights ≈ 49.6 m and 29.6 m

Let the shorter pillar have height h m, so the higher one is (h + 20) m. Let the point be x m from the foot of the higher pillar, so (80 – x) m from the other.
Higher pillar: $tan 6 0^{\circ} = \frac{h + 20}{x}$ , so $h + 20 = 3 x$ .
Other pillar: $tan 3 0^{\circ} = \frac{h}{80 - x}$ , so $h = \frac{80 - x}{3}$ .
$3 x - 20 = \frac{80 - x}{3}$ gives $3 x - 203 = 80 - x$ , so $4 x = 80 + 203$ , $x = 20 + 53$ .
x = 20 + 5(1.73) = 28.65 m; distance from the other pillar = 80 – 28.65 = 51.35 m.
Higher pillar $= 3 x = 203 + 15 = 34.6 + 15 = 49.6$ m.
Other pillar = 49.6 – 20 = 29.6 m.

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