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**Short Answer type Questions**

**Q 1. The ratio of the height of a tower and the length of its shadow on the ground is √ 3 : 1. What is the angle of elevation of the sun? [Delhi 2017]****Ans:**

Let, AB = height of the tower

BC = length of the shadow

∠ACB = θ = Angle of elevation of the sum

In ∠ABC, ∠B = 90^{0}**Q 2. An observer, 1.5 m tall, is 28.5 m away from a 30 m high tower. Determine the angle of elevation of the top of the tower from the eye of the observer. [Delhi 2017 (C)]****Ans: **Draw AM || BD

Here, AB - DM - 1 . 5 m

CM = CD - DM

= 30 - 1.5 = 28.5 m

Let θ be the angle of elevation of the top of the tower from the eye of the observer

∴ In ΔACM**Q 3. If a tower 30 m high, casts a shadow 10√ 3 m long on the ground, then what is the angle of elevation of the sun [AI 2017]Ans:**

Let θ be the angle of elevation of the sun

∴ In right angled triangle ABC

Hence, the angle of elevation of the sun is 60

Ans:

Observe that AD is a transversal to the parallel lines AE and BD. Therefore ∠EAD and ∠ADB are alternate angles and so both are equal

So, ∠ADB = 45°

Similarly, ∠ACB = 60°

Let AB be the cliff.

∴ AB = 150 m

After 2 minutes it reaches at D.

In right triangle ABC,

In right triangle ABD,

Distance covered in 2 minutes

Hence, speed of boat is 31.7 m/min**Q 5. A ladder 15 m long makes an angle of 60° with the wall. Find the height of the point where the ladder touches the wall. [CBSE(F) 2017]Ans:**

Ans:

Let the speed of boat be x m/min

∴ CD = 2x

In ΔABC

In ΔABD

(∵ y = 50√3)

⇒ x = 25(3-√3)

speed = 25(3-√3)m/min

∴ = 25 x 60 (3-√3)m/h = 1500(3-√3)m/h

Ans:

DB = (6 - 2.54)m = 3.46 m

⇒

⇒ x = 20m

∴ AB = 20 + 1.7 = 21.7m

Let height of tower be x m an d distance of point from tower be y m.

(i) from the fig

⇒ x + 5 = 3x ⇒ x = 5/2 = 2.5

Height of tower = 2.5 m

Distance of point from tower = y = √3x

= (2.5 x 1.732) or 4.33 m

Ans:

ln ΔABM_{1},

Let AB be the building having height 75 m and the angles of elevation are 30° and 60° from the point M_{1} and M_{2} respectively;

In ΔABM_{2}

⇒

∴ M_{1}M_{2} = M_{1}B + BM_{2}

= 75√3 + 25√3 = 100√3m = 173m

∴ Distance between two men = 173 m.

**Long Answer Type Questions**

**Q 1. Two poles of equal heights are standing opposite to 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 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles****[NCERT; CBSE 2019 (30/1/2)]****Ans:**

∠APB = 60° and ∠DPC = 30°

The distance between two poles is 80m. (Given)

Let AP = x m, then PC = (80 - x) m

Now, in ΔAPB, we have

⇒ ...(i)

Again in ΔCPD, we have

...(ii)

From (i) and (ii), we have

Now, putting the value of x in equation (i), we have

Hence , the height of the pole is 20√3 m and the distance of the point from first pole is 20m and that of the second pole is 60 m.**Q 2. As observed from the top of a 100 m high light house from the sea-level, the angles of depression of two ships are 30° and 45°. 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] [CBSE 2018]****Ans: **Let AB be the height of the light house from the sea-level. C and D are the two ships. One ship is exactly behind each other such that

∠XAC = ∠ACB

= 30° and ∠XAD

= ∠ADB = 45°

We have to find distance between two ships, i.e. CD

In ΔABC, ∠B - 90°

Also, in Δ ABD, ∠B = 90°

From (i), we have**Q 3. The angle of elevation of the top of a hill at the foot of a tower is 60° and the angle of depression from the top of tower to the foot of hill is 30°, If tower is 50 metre high, find the height of the hill. [Delhi 2017, CBSE 2018 (C)]Ans:**

Equating (i) and (ii), we get

Put DC = 50 m

AB = 3 x 50 = 150 m

Height of the hill = 150 m

Ans:

Let A be the position of the aeroplane and let C and D be two points on the two banks of the river such that the angles of depression at C and D are 60° and 45° respectively.

Let BC = x metres

and BD = y metres

We have to find CD i.e. (x + y)m

In right angled triangle ABC.

In right angled triangle ABD,

From equations (i) and (ii), we get

Hence, the width of the river is 473.2 metres.

Ans:

[From (i)]

30 = h + 10

h = 20 m

The height of helicopter above the ground is 20 m.

Amit and Deepak are on opposite sides of the bird

Sol: In ΔACB,

Let bird is at A and after 2 seconds it reaches at E.

∴ Distance covered = AE.

In right ΔABC,

In right ΔEDC,

**Q 8. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower. [CBSE Delhi 2017]****OR****From the top of a 7 m high building, the angle of elevation of the top of a tower is 60° and the angle of depression of its foot is 45°. Find the height of the tower. [Use √3 = 1.732] [CBSE (F) 2017]****Ans: **

Let PQ be the building of height 7 metres and AB be the cable tower. Now it is given that the angle of elevation of the top A of the tower observed from the top P of building is 60° and the angle of depression of the base B of the tower observed from P is 45°.

So, ∠APR = 60° and ∠QBP = 45°

Let QB = xm , AR = h m, then , PR = x m

Now, in ΔAPR, we have

⇒ √3x = h ⇒ h = √3x ..(i)

Again, in ΔPBQ we have

⇒ x = 7 ..(ii)

Putting the value of x in equation (i), we have

So, the height of tower = AB = AR + RB = 7√3 + 7 = 7(√3+1)m.**Q 9. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point. [CBSE Delhi 2017 (C)]Ans:**

Let OA be the tower of height h, and P be the initial position of the car when the angle of depression is 30°.

After 6 seconds, the car reaches to Q such that the angle of depression at Q is 60°. Let the speed of the car be v metre per second. Then,

PQ = 6v (∵ Distance = speed x time)

and let the car take t seconds to reach the tower OA from Q as shown in figure. Then, OQ = vt metres. Now, in ΔAQO, we have

... (i)

Now, in ΔAPO, we have

Now, substituting the value of h from (i) into (ii), we have

Hence, the car will reach die tower from Q in 3 seconds.**Q 10. From the top of a tower, 100 m high, a man observes two cars on the opposite sides of the tower and in same straight line with its base, with angles of depression 30° and 45°. Find the distance between the cars. [Take S = 1.732] [CBSE (AI) 2017]Ans:** Let AO be the tower of height 100 m. Car B and Car C are in opposite direction and at distance of jc m and y m respectively.

In ΔABO,

⇒ x = 100

In ΔACO,

⇒ y = 100√3

Distance between the cars - x + y

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