# Height and Distance - Aptitude Questions and Answers - RejinpaulPlacement

15.
On the same side of a tower, two objects are located. Observed from the top of the tower, their angles of depression are 45° and 60°. If the height of the tower is 150 m, the distance between the objects is:

Solution:   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 = (150 - 150/√3 ) m = ( (150 (√3 - 1))/√3 * √3/√3 ) m = 50 (3 - √3 ) m

CD = (50 * 1.27) m = 63.5 m
16.
A man on the top of a vertical observation tower observes a car moving at a uniform speed coming directly towards it. If it takes 12 minutes for the angle of depression to change from 30° to 45°, how soon after this will the car reach the observation tower?

Solution:   Let AB be the tower and C and D be the two positions of the car.

Then, ∠ACB = 45°, ∠ADB = 30°.

Let AB = h, CD = x and AC = y.

AB/AC = tan 45° = 1

h/y = 1 => y = h

AB/AD = tan 30° = 1/√3

=> h/(x + y) = 1/√3 => x + y = √3 h

x = (x + y) - y = √3 h - 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)) min = 1200/73 min = 16 min. 23 sec.

17.
The top of a 15 metre high tower makes an angle of elevation of 60° with the bottom of an electric pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?

Solution:   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 * √3 ) = √3 (15 - h)

AC = DE => 15/√3 = √3 (15 - h)

=> 3h = (45 - 15)

=> h = 10 m

AC = DE => 15/√3 = √3 (15 - h)

=> 3h = (45 - 15)

=> h = 10 m