Question 1

At a point, 15 m away from the base of a 15 m high house, the angle of elevation of the top is :

Accepted Answer

Question 2

A tower subtends an angle of 30° at a point on the same level as the foot of the tower. At a second point, h m above the first, the depression of the foot of the tower is 60°. The horizontal distance of the tower from the point is :

Accepted Answer

Question 3

The angle of elevation of the sun when the length of the shadow of a pole is √3 times the height of the pole will be :

Accepted Answer

Question 4

The tops of two poles of height 20 m and 14 m are connected by a wire. If the wire makes an angle 30° with the horizontal, then the length of the wire is :

Accepted Answer

Question 5

The top of a 15 m high tower makes an angle of elevation of 60° with the bottom of an electric pole and an angle of elevation of 30° with the top of the pole. What is the height of the electric pole?

Accepted Answer

Question 6

From a 15 m high bridge on the river, the angle of depression of a boat is 30°. If the speed of the boat be 6 Km/h, then the time taken by the boat to reach exactly below the bridge will be :

Accepted Answer

Question 7

The angle of elevation of a tower from a distance 50 m from its foot is 30°. The height of the tower is :

Accepted Answer

Question 8

If r sin θ = 1, r cos θ = √ 3 then the value of ( √ 3 tan θ + 1 ) is :

Accepted Answer

Question 9

Two posts are x meter apart and the height of one is double that of the other. If from the midpoint of the line joining their feet, an observer finds the angular elevations of their tops to be complementary, then the height (in meter) of the shorter…

Accepted Answer

Question 10

The angles of depression of two ships from the top of a light house are 45° and 30° towards east. if the ships are 200 m apart, find the height of the light house .

Accepted Answer

Let us draw a figure below as per given question. Let AB = h meter be the height of the tower and two ships are situated at D and C respectively; such that, CD = 200 m; ∠ADC = 30° ∠ACB = 45° and BC = x meter (say) Now from right triangle ABC, tan 45° = h/x ⇒ 1 = h/x ∴ x = h Again from right triangle ABD, tan 30° = h/( 200 + x ) ⇒ 1/√3 = h/( 200 + x ) Since x = h , we will get. ⇒ 1/√3 = h/(200 + h ) ⇒ (200 + h ) = h X √3 ⇒ h X √3 - h = 200 ⇒ h x 1.732 - h = 200 ⇒ h(1.732 - 1) = 200 ∴ h = 200/0.732 = 273.2 m = 273 m

Question 11

At a point on a level plane a tower subtends an angle Θ and a flag-staff a ft. in length at the top of the tower subtends an angle Θ . The height of the tower is :

Accepted Answer

Let us draw a figure as per given question. Let OP be the tower of height h (say) and PQ the flag-staff of height a, such that ∠OAP = Θ and ∠PAQ = Φ In ΔOAP use the formula TanΘ = P/B = Perpendicular distance / Base distance ⇒ TanΘ = OP/OA ⇒ OA = OP CotΘ Put the value of OP in above equation, we will get ⇒ OA = hCotΘ ................ (1) Now from triangle ΔOAQ, ⇒ Tan(Θ + Φ) = OQ/OA ⇒ OA = OQ Cot(Θ + Φ) ⇒ OA = (h + a) Cot(Θ + Φ) .............(2) from equation (1) and (2), We will get ∴ hCotΘ = (h + a) Cot(Θ + Φ) ⇒ CotΘ / Cot(Θ + Φ) = (h + a) / h ⇒ CotΘ / Cot(Θ + Φ) = h / h + a / h ⇒ CotΘ / Cot(Θ + Φ) = 1 + a / h ⇒ (CotΘ / Cot(Θ + Φ)) - 1 = a / h ⇒ (CotΘ - Cot(Θ + Φ) ) / Cot(Θ + Φ) = a / h ⇒ h = aCot (Θ + Φ) / CotΘ - Cot(Θ + Φ) After converting Cot into Sin and Cos, We will get ⇒ h = aSinΘCos(Θ + Φ)/SinΦ

Question 12

A 6 ft-tall man finds that the angle of elevation of the top of a 24 ft-high pillar and the angle of depression its base are complementary angles. Then the distance between pillar and man is?

Accepted Answer

Let us draw a figure below from the given question. Let AB = 24 ft and CD = 6 ft be the height of the pillar and man respectively. Here ∠ ACE = ∝ ∠ CBD = 90° - ∝ Now from right triangle CDB, BD = 6 cot (90° - ∝) = 6 tan ∝ ∴ tan ∝ = BD/6 ...............(i) From right triangle ACE, tan ∝ =18/EC = 18/BD ...............(ii) (∵ EC = BD) Now from (i) and (ii) we get, BD/6 = 18/BD ⇒ BD2 = 18 x 6 ∴ BD = 6√3 ft. Hence, distance between pillar and man = 6√3 ft

Question 13

Two ships leave a port at the same instant. One sails at 30km/hr in the direction N 32° E while the other sails at 20 km/hr in the direction S 58° E. After two hours the ships are distant from each other by.

Accepted Answer

Let us draw a figure below as per given question. Let two ships started from point O at the speed of 30 km/hr and 20 km/hr respectively, after two hours they reach at points A and B. Now, ∠NOA = 32° and ∠SOB = 58° , Then, ∠AOB = 180° - (32° + 58°) = 90° Since ΔAOB is a right triangle in which OA = 2 x 30 = 60 km and OB = 2 x 20 = 40 km Since, AB = √OA2 + OB2 = √(60)2 + (40)2 = √5200 = 20√13

Question 14

ABC is a triangular park with AB = AC = 100 meters. A clock tower is situated at the mid-point of BC. The angles of elevation of the top of the tower of A and B are cot -1 (3.2) and cosec -1(2.6) respectively. The height of the tower in meters is?

Accepted Answer

Let us draw a figure below as per given question. Given ∝ = cot-1 (3.2) and β = cosec-1(2.6) In triangle ΔPAD, AD = h cot ∝ In triangle ΔPBD, BD = h cot β In triangle Δ ABD, AB2 = AD2 + BD2 = h2(cot2∝ + cot2 β ) ⇒ 1002 = h2{ cot2 ∝ + (cosec2 β - 1) } ⇒ 1002= h2 { (3.2)2 + (2.6)2 - 1} = 16h2 ⇒ h = 25 m.

Question 15

ABCD is a square plot. The angle of elevation of the top of a pole standing at D from A or C is 30° and that from B is Θ then Tan Θ equal to?

Accepted Answer

Let a be the length of a side of square plot ABCD and h, the height of the pole standing at D. Since elevations of P from A or C is 30° and that from B is Θ, ∴ In triangle Δ PCD, tan 30° = h/a i.e h/a = 1/√3 .............. (1) And in triangle Δ PBD, Since PD = h and BD = √(AB2 + AD)2 = a√2. Put the value of PD and BD , we will get tan Θ = PD/BD = h/(a√2) put the value of h/a from equation (1) tan Θ = 1/√2 x 1/√3 tan Θ = 1/√6

Question 16

The height of the center of the round balloon of radius r, which subtend an angle ∝ at the eye of an observer and the elevation of whose center from the eye is β, is given by?

Accepted Answer

Let O be the centre of the balloon of radius r which subtend an angle ∝ at the eye of an observer at E . If EA and EB are the tangents to the ballon, then ∠ OEA = ∠ OEB = ∝/2 In triangle ΔOAE, Sin ∝/2 = OA/OE ∴ OE = r cosec 1/2 ∝ In ∠OEL, height of the center of the balloon = h = OE sin β = r Cosec ∝/2 Sin β.

Question 17

At a point on a level plane a tower subtends an angle θ and a flag-staff a ft. in length at the top of the tower subtends an angle ɸ. The height of the tower is :

Accepted Answer

Let OP be the tower of height h (say) and PQ the flag-staff of height a, such that ∠OAP = θ and ∠PAQ = ɸ In ΔOAP and ∠OAQ OA = OP cot θ = h cotθ and OA = OQ cot ( θ + ɸ ) = (h + a) cot ( θ + ɸ ) ∴ h cotθ = (h + a) cot ( θ + ɸ ) ⇒ h = a cot ( θ + ɸ ) / cotθ - cot ( θ + ɸ )

Question 18

A tower stands on a horizontal plane. A man on the ground 100 m from the base of the tower finds the angle of elevation of the top of the tower to be 30°. What is the height of the tower?

Accepted Answer

Question 19

The upper part of a tree broken by the wind makes an angle of 30° with the ground and the distance from the root to the point where the top of the tree touches the ground is 10 m. What was the height of the tree?

Accepted Answer

Question 20

In a rectangle, if the angle between a diagonal and a side is 30° and the length of diagonal is 6 cm, the area of the rectangle is :

Accepted Answer

Height and Distance Questions