Arithmatical  Heights and Distance
DIRECTIONS : Problems based on Heights and Distance.
1. 
A ladder learning against a wall makes an angle of 60° with the ground. If the length of the ladder is 19 m, find the distance of the foot of the ladder from the wall. 

A. 9 m 

B. 9.5 m 

C. 10.5 m 

D. 12 m 
Solution 

Let AB be the wall and BC be the ladder. 
Then, < ABC  = 60° 
and, BC  = 19 m.; 
AC = x metres 
AC/BC  = cos 60° 
= x / 19 
=1 / 2 
x= 19/2 
= 9.5 m. 

2. 
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 

A. 2.3 m 

B. 4.6 m 

C. 7.8 m 

D. 9.2 m 
Solution 
Let AB be the wall and BC be the ladder. 

Then, < ABC  = 60° 
AC  = 4.6 m.; 
AC/BC  = cos 60° 
= 1 /2 
‹=›BC=2×AC 
=(2× 4.6) m 
= 9.2 m 

3. 
The angle of elevation of the sun, when the length of the shadow of a tree is √3 times the height of the tree is 

A. 30° 

B. 45° 

C. 60° 

D. 90° 
Solution 
Let AB be the tree and AC be its shadow. 

Then, < ABC  = θ. 
Then, AC/AB  = √3 
cotθ= √3 
θ=30° 

4. 
If the height of a pole is 2√3 metres and the length of its shadow is 2 metres, find the angle of elevation of the sun. 

A. 30° 

B. 45° 

C. 60° 

D. 90° 
Solution 
Let AB be the pole and AC be its shadow. 

Then, < ACB  = θ. 
Then, AB= 2 √3 m, AC = 2m, 
tanθ= AB/AC=2√3/2 
= √3 
θ=60° 

5. 
From a point P on a level ground, the angle of elevation of the top tower is 30°. If the tower is 100 m high, the distance of point P from the foot of the tower is: 

A. 149 m 

B. 156 m 

C. 173 m 

D. 200 m 
Solution 
Let AB be the tower. 

Then, <APB =30° and AB= 100 m 
AB/AP= tan30° 
= 1 / √3 
AB = (AB x √3)= 100√3 m. 
= (100 x 1.73) m = 173 m. 
