Question 1

A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30° with the man's eye. The man walks some distance towards the tower to watch its top and the angle of the elevation becomes 60°. What is the distance …

Accepted Answer

One of AB, AD and CD must have given. So, the data is inadequate.

Question 2

The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is:

Accepted Answer

Let AB be the wall and BC be the ladder. Then, ∠ACB = 60° and AC = 4.6 m. AC = cos 60° = 1 BC 2 ⟹ BC = 2 x AC = (2 x 4.6) m = 9.2 m.

Question 3

An observer 1.6 m tall is 20√3 away from a tower. The angle of elevation from his eye to the top of the tower is 30°. The heights of the tower is:

Accepted Answer

Let AB be the observer and CD be the tower. Draw BE ⊥ CD. Then, CE = AB = 1.6 m, BE = AC = 20√3 m. DE = tan 30° = 1 BE √3 ⟹ DE = 20√3 m = 20 m. √3 ∴ CD = CE + DE = (1.6 + 20) m = 21.6 m.

Question 4

From a point P on a level ground, the angle of elevation of the top tower is 30°. If the tower is 100 m high, the distance of point P from the foot of the tower is:

Accepted Answer

Let AB be the tower. Then, ∠APB = 30° and AB = 100 m. AB = tan 30° = 1 AP √3 ⟹ AP = (AB x √3) m = 100√3 m = (100 x 1.73) m = 173 m.

Question 5

The angle of elevation of the sun, when the length of the shadow of a tree √3 times the height of the tree, is:

Accepted Answer

Let AB be the tree and AC be its shadow. Let ∠ACB = Θ. Then, AC = √3 ⟹ cot Θ = √3 AB ∴ Θ = 30°.

Question 6

Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is:

Accepted Answer

Let AB be the lighthouse and C and D be the positions of the ships. Then, AB = 100 m, ∠ACB = 30° and ∠ADB = 45°. AB = tan 30° = 1 ⟹ AC = AB x √3 = 100√3 m. AC √3 AB = tan 45° = 1 ⟹ AD = AB = 100 m. AD ∴ CD = (AC + AD) = (100√3 + 100) m = 100(√3 + 1) = (100 x 2.73) m = 273 m.

Question 7

The angle of elevation of a ladder leaning against a wall is 60º and the foot of the ladder is 12.4 m away from the wall. The length of the ladder is :

Accepted Answer

Consider the diagram is shown above where PR represents the ladder and RQ represent the wall. Cos 60° = PQ / PR ⇒ 1 / 2 = 12.4 / PR ⇒ PR = 2 × 12.4 = 24.8 m

Question 8

A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is 60°, when he retires 40 meters from the bank then he find the angle to be 30°. Then the breadth of the river is?

Accepted Answer

Let us draw a figure below as per given question. Let A be the position of a person on the bank of a river and OP the tree on the opposite bank and ∠OAP = 60°. When the person retires to the position B, then AB = 40 meter and ∠OBP = 30° Let us assume OA(Breadth of the river) = x meter and height of tree OP = h meter In ΔOAP, Use the trigonometry formula Tan60° = P/B = Perpendicular distance / Base distance ⇒ Tan60° = OP / OA ⇒ OP = OA Tan60° Put the value of OP and OA, We will get ⇒ h = x√3 ..............(1) Now in the triangle ΔOBP Tan30° = OP / OB ⇒ OP = OB Tan30° ⇒ OP = (x + 40)/√3 ⇒ h = (x + 40)/√3 ...................(2) From Equation (1) and (2), We will get ⇒ (x + 40)/√3 = x√3 ⇒ (x + 40) = x√3 X √3 ⇒ (x + 40) = 3x ⇒ 3x - x = 40 ∴ x = 20 m

Question 9

A man is observing from the top of a tower a boat speeding away from the tower. The boat makes an angle of depression of 45° with the man's eye when at a distance of 60 m from the tower. After 5 second, the angle of depression becomes 30°, Find the …

Accepted Answer

Let us draw the figure from the given question. Let AB = h meter be the height of the tower; C and D are the two points on the ground such that BC = 60 m; ∠ACB = 45° and ∠ADB = 30° Now from right triangle ABC, tan 45° = h/60 ⇒ 1 = h/60 ∴ h = 60 m; Again from right triangle ABD; tan 30° = h/(x + 60) ⇒ 1/√3 = 60/(x + 60) ⇒ x + 60 = 60√3 ∴ x = 60(1.73 - 1) = 43.8 meter Hence, speed of boat = 43.8/5 m/s = 43.8/5 x 18/5 = 31.5 km/hr.

Question 10

From a point P on a level ground, the angle of elevation of the top of a tower is 30°, if the tower is 100 m high, find the distance of point P from the foot of the tower.

Accepted Answer

Let us draw a figure from given question. Let, AB = 100 m be the height of a tower, P is a point on the ground such that ∠APB = 30° From right triangle ABP, BP = 100 cot 30° ⇒ BP = 100 √3 ⇒ BP = 100 X 1.73 ⇒ BP = 173 meter

Question 11

A vertical pole is 75m high. Find the angle subtended by the pole a point 75 m away from its base.

Accepted Answer

Let us draw a figure below as per given question. Let AB = 75 m be the height of pole and C is a point on the ground such that BC = 75m Now, from right triangle ABC. tan ∝ = AB/BC = 75/75 ⇒ tan∝ = 1 ∴ ∝ = 45°

Question 12

A person standing on the bank of river finds that the angle of elevation of the top of a tower on opposite side bank is 45°. Which of the following statement is correct

Accepted Answer

Let us draw a figure below as per given question. Let AB = h m be the height of the tower, BC = x m be the breadth of the river and also ∠ACB = 45° Now from right triangle ABC tan 45° = h/x ⇒ 1 = h/x ∴ x = h Hence, breadth of the river = height of the tower

Question 13

The angle of elevation of the top of two vertical towers as seen from the middle point of the line joining the foot of the towers are 60° and 30° respectively. The ratio of the heights of the towers is?

Accepted Answer

Let AB and CD be two towers of height h1 and h2 respectively and O the mid-point of the line joining the foots A and C of the towers. Let OA = OC = x Then h1 = x tan 60° = x√3 and h2 = x tan 30° = x/√3 ∴ h1 /h2= 3/1. Hence, h1 : h2 = 3 : 1

Question 14

On the ground level the angle of elevation of the top of a tower is 30°, On moving 20 m nearer the tower, the angle of elevation found to be 60°. The height of the tower is?

Accepted Answer

Let us draw a figure below from the given question. Let AB = h meter be the height of the towers. B and C are two points such that BC = 20 m; ∠ADB = 30° and ∠ACB = 60° BC = x meter (let us assume) Now, from right triangle ABC, x = h cot 60° ⇒ x = h/√3 meter Again, from right triangle ABD, h = (20 + x) tan 30° put the value of x in above equation. ∵ x = h/√3 ⇒ h = (20 + h/√3) x 1/√3 ( ∵ tan 30° = 1/√3 ) ⇒ h - h/3 = 20/√3 ⇒ 2h/3 = 20/√3 ∴ h = 20 x 3/2√3 = 10√3 m

Question 15

The ratio of the length of a rod and its shadow is 1: √3 .The angle of elevation of the sun is :

Accepted Answer

Question 16

When the sun is 30° above the horizontal, the length of shadow cast by a building 50 m high is :

Accepted Answer

Question 17

A tower stands at the end of a straight road. The angles of elevation of the top of the tower from two points on the road 500 m apart are 45° and 60°, respectively. Find out the height of the tower.

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Question 18

The height of a tower is 100 m. When the angle of elevation of the sun changes from 30° to 45°, the shadow of the tower becomes P meter smaller. The value of P is :

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Question 19

From the top of a 25 m high, cliff the angle of elevation of a tower is found to be equal to the angle of depression of the foot of the tower. Find out the height of the tower.

Accepted Answer

Question 20

The angle of elevation of the top of a tower from a point 20 m away from its base is 45°. The height of the tower is :

Accepted Answer

Height and Distance Questions