Height and Distance with Explanation

Q1. A ladder of length 19m is resting against a wall. It makes an angle of 60° with the ground. What is the distance of the ladder from the wall?

Answer- Option D

Explanation-

Let AB be the wall and BC be the ladder.
Then, ∠ ACB = 60° and BC = 19 m
Let AC = x meters

AC BC

= cos60°

x 19

=

1 2

x

=

19 2

= 9.5

Distance of the foot of the ladder from the wall = 9.5 m

Q2. Gaurav is standing on the bank of river observes that the angle suspended by a tree on the opposite bank is 60°. When he goes 36 m away from the bank, he finds the angle to be 30° . What is the width of the river?

Answer- Option A

Explanation-

Let AB be the tree and AC be the river. Let C and D be the two positions of Gaurav.
Then, ∠ ACB = 60° , ∠ ADB = 30° and CD = 36 m
Let AB = h meters and AC = x meters

AB AD

= tan30°

=

1   3

h 36 + x

=

1   3

h

=

36 + x   3

...(i)

AB AC

= tan60°

=   3

h x

=   3

h =   3x ...(ii)
From (i) and (ii), we get :

36 + x   3

=

3x

x = 18m

Q3. The length of the shadow of a tower is 2 meter, and its height is 2  3 meters. Find the angle of elevation of the Sun?

Answer- Option B

Explanation-

Let AB be the tower and AC be its shadow.
Let angle of elevation, ∠ ACB = Θ
Then, AB = 2  3 m, AC = 2 m.

tan Θ

=

AB AC

=

2  3 2

=   3

Θ = 60°
So, the angle of elevation is 60°

Q4. To objects are situated on the same of a building. Observed from the top of the building their angles of depression are 45°, and 60°. Find the distance between the objects, if the height of the building is 150 meter?

Answer- Option C

Explanation-

Let AB be the tower and C and D be the objects
Then, AB = 150 m, ∠ ACB = 45° and ∠ ADB = 60°

AB AD

=

tan 60°

=   3

AD =

AB   3

=

150   3

m

AB AC

=

tan 45°

= 1

AC = AB = 150 m

CD = (AC - AD)

150 -

150   3

m

= 63.5 m

Q5. The top of a tree makes an angle of elevation of 30° with the top of a pillar and 60° with the bottom of a pillar. If the height of the tree is 15 meter, find the height of the pillar?

Answer- Option D

Explanation-

Let AB be the tower and CD be the electric pole.
Then, ∠ ACB = 60° , ∠ EDB = 30° and AB = 15 m
Let CD = h Then, BE = (AB - AE) = (AB - CD) = (15 - h)

AB AC

=

tan 60°

=   3

AC =

AB   3

=

15   3

And,

BE DE

=

tan 30°

=

1   3

DE = (BE x   3)

=   3(15 - h)

AC = DE

15   3

=   3(15 - h)

3h = (45 - 15) h = 10m

Q6. Two temples opposite to each other are located on each bank of a river Ganga. One temple is 54 meter height. From the top of this temple the angle of depression of the foot and the top of other temple are 60° and 30° respectively. Find the height of the other temple and the breadth of the river Ganga?

Answer- Option C

Explanation-

Let AB and CD be the two temples and AC be the river.
Then, AB = 54 m
Let AC = x meters and CD = h meters
∠ ACB = 60°, ∠ EDB = 30°

AB AC

= tan 60° =   3

AC =

AB   3

=

54   3

=

54   3

*

3   3

= 18  3 m

DE = AC = 18   3 m

BE DE

= tan 30° =

1   3

BE =

18  3

*

1   3

= 18 m

CD = AE = AB - BE = (54 - 18) m = 36 m
So, width of the river Ganga = AC = 18  3 m = (18 * 1.73) m = 31.14 m
Height of the other temple = CD = 18 m

Q7. A bike is moving at a uniform speed towards an observation tower. It takes 12 minutes for the angle of depression to change from 30° to 45°. After this, how much time will this bike take to reach the observation tower?

Answer- Option D

Explanation-

Let AB be the observation tower and C and D be the two positions of the bike.
Then, ∠ ACB = 45°, ∠ ADB = 30°
Let AB = h, CD = x and CD = y

AB AC

= tan 45° = 1

h y

= 1

y = h

AB AD

= tan 30° =

1   3

h x + y

=

1   3

x + y =   3h
x = (x + y) - y =   3h - h = h(  3 - 1)
Now, h(  3 - 1) is covered in 12 min.
So, h will be covered in

12 h(  3 - 1)

* h

=

12 (  3 - 1)

Ans is 16 min 23 sec