Q: 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

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

Q: From the top of h meter high cliff the angles of depression of the top and the button of a tower are observed to be 30° and 60° respectively. The height of the tower is?

Let us draw a figure below as per given question. Let AB = h meter be the height of cliff and CD = x meter be the height of the tower and also ∠ ADB = 60° and ∠ACE = 30° , Now, from the figure, AE = (h - x) meter From right triangle ABD, BD = h cot 60° = h/√3 m From right triangle CEA. (h - x) = EC tan 30° = BD tan 30° ⇒ h - x = h/√3 X 1/√3 ∴ x = h - h/3 = 2h/3 meter

Question 1

The angle of elevation of the top of a hill from each of the vertices A, B, C of a horizontal triangle is α. The height of the hill is :

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Question 2

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 m from the bank, he finds the angle to be 30°. The breadth of the river is :

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Question 3

The angle of elevation of the top of an unfinished tower at a point distant 120 m from its base is 45°. If the elevation of the top at the same point is to be 60°, the tower must be raised to a height :

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Question 4

The angle of elevation of the top of a tower at a point G on the ground is 30°. On walking 20 m towards the tower, the angle of elevation becomes 60°. The height of the tower is equal to :

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Question 5

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

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Question 6

A tower standing on a horizontal plane subtends a certain angle at a point 160 m apart from the foot of the tower. On advancing 100 m towards it, the tower is found to subtend an angle twice as before. The height of the tower is :

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Question 7

A man from the top of a 100 m high tower sees a car moving towards the tower at an angle of depression of 30°. After some time, the angle of depression becomes 60°. The distance (in m) traveled by the car during this time is :

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Question 8

sin A + sin A is ( 0° < A < 90° ) . 1 + cos A 1 - cos A

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Question 9

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

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

Question 10

From the top of h meter high cliff the angles of depression of the top and the button of a tower are observed to be 30° and 60° respectively. The height of the tower is?

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Let us draw a figure below as per given question. Let AB = h meter be the height of cliff and CD = x meter be the height of the tower and also ∠ ADB = 60° and ∠ACE = 30° , Now, from the figure, AE = (h - x) meter From right triangle ABD, BD = h cot 60° = h/√3 m From right triangle CEA. (h - x) = EC tan 30° = BD tan 30° ⇒ h - x = h/√3 X 1/√3 ∴ x = h - h/3 = 2h/3 meter

Question 11

The angles of elevation of the top of a verticle tower from two points, distance a and b (a > b) from the base and in the same straight line with it are complementary. Then the height of the tower is?

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Let CD = h unit be the height of the tower and A and B be the two points on the ground, such that DA = a; DB = b; ∠ DAC = ∝ and ∠DBC = 90° - ∝ From right triangle ADC, CD = h = a tan ∝ ...(i) From right triangle BDC, CD = h = b tan (90° - ∝ ) = b cot ∝ .......(ii) Multiplying equations (i) and (ii), we get h2 = a tan ∝ X b cot∝ Hence, h = √ab

Question 12

The angle of elevation of the top of a tower standing on a horizontal plane from a point A is α. After walking a distance a towards the foot of the tower, the angle of elevation is found to be β. The height of the tower is :

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Let OP be the tower of height h (say) and A and B be the two positions on the horizontal line through O, such that ∠OAP = α, ∠OBP = β and OB = x In ΔOBP, Use the trigonometry formula Tanβ = P/B = Perpendicular distance / Base distance Tanβ = OP/OB ⇒ OB = OP/Tanβ ⇒ OB = OP Cotβ Put the value of OB and OP , We will get x = h Cot β...............(1) In ΔOAP, Similarly Tanα = OP/OA ⇒ OA = OP/ Tanα ⇒ OA = OP Cot α Put the value of OA and OP ⇒ a + x = h Cot α ⇒ x = h Cot α - a ............(2) From equation (1) and (2) ∴ h Cot β = h Cot α - a ⇒ a = h Cot α - h Cot β ⇒ a = h (Cot α - Cot β) ⇒ a = h (Cos α/ Sin α - Cos β / Sin β ) ⇒ a = h( (Cos α Sin β - Cos β Sin α ) /Sin α Sin β ) ⇒ a = h( Sin(β - α) / Sin α Sin β) ⇒ h = a Sin α Sin β/ Sin(β - α)

Question 13

A man on the top of a vertical towers observes a car moving at a uniform speed coming directly towards it. if it takes 12 minute for the angle of depression to change from 30° to 45°, how soon after this will the car reach the tower?

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Let us draw a figure below from given question. Let AB = h meter be the height of the tower B and C are two points such that ∠ACB = 30° ∠ADB = 45° and CD = x meter (say) From right triangle ABD, tan 45° = h/BD ∴ BD = h meter; Again from right triangle ABC tan 30° = h/(h + x ) ⇒ h + x = √3 h ∴ x = (1.73 - 1)h = 0.73h Now, 0.73h meter covered in 12 min Hence, h meter covered in 12/0.73 = 1200/73 min = 16 min 23 sec .

Question 14

An observer measures angles of elevation of two tower of equal height from a point between the towers. If the angles of elevation are 60° and 30° and distance of nearer tower is 100 m then the height of each tower and the distance between the towers…

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Let us draw a figure below as per given question. Let AB = CD = h meter be the heights of the towers. E is a point such that DE = 100 meter; ∠CED = 60° and ∠AEB = 30° Now, BE = x meter (say) From right triangle CDE. h = 100 tan 60° ⇒ h = 100√3 meter From right triangle ABE, x = h cot 30° put the value of h, we will get x = 100√3 X √3 x = 100 X 3 = 300 meters Distance between the tower = DE + EB = 100 + 300 = 400 meters Height of the tower = h = 100√3 meter

Question 15

A man 2 m high, walks at a uniform speed of 6 m/min away from a lamp post, 5 m high. Find the rate at which the length of his shadow increases .

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Let us draw a figure as per given question. Let AB = 5 meter and CD = 2 meter be the heights of a lamp post and the men respectively. At any time t BD = x meter and shadow of man ED = y meter. Then, dx/dt = 6 m/min Now, right triangles ABE and CDE are similar, then AB/CD = BE/DE ⇒ 5/2 = (x + y)/y ⇒ 5y = 2x + 2y ⇒ 5y - 2y = 2x ⇒ 3y = 2x ⇒ 3 dy/dt = 2 dx/dt ⇒ 3 dy/dt = 2 x 6 ⇒ dy/dt = 2 x 6 / 3 ∴ dy/dt = 2 x 2 = 4 Hence, length of his shadow increase at the rate of 4 m/min.

Question 16

The angle of elevation of moon when the length of the shadow of a pole is equal to its height, is:

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Question 17

A pole being broken by the wind, the top struck the ground at an angle of 30° and at a distance of 21 m from the foot of the pole. Find out the total height of the pole.

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Question 18

The shadow of a tower standing on a level plane is found to be 50 m longer when the sun's altitude is 30° than when it is 60°. Find the height of the tower.

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Question 19

A 20 m high electric pole stands upright on the ground with the help of steel wire to its top and affixed on the ground. If the steel wire makes 60° with the horizontal ground, then find out the length of the steel wire.

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Question 20

When the length of the shadow of a pole is equal to the height of the pole, then the elevation of source of light is :

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Height and Distance Questions