Question

A tree (TS) of height 30 m stands in front of a tall building (AB). Two friends Rohit and Neha are standing at R and N respectively, along the same straight line joining the tree and the building (as shown in the diagram). Rohit, standing at a distance of 150 m from the foot of the building, observes the angle of elevation of the top of the building as 30°. Neha from her position observes that the top of the building and the tree has the same elevation of 60°. Find the:

1. height of the building 
2. distance between 
   1. Neha and the foot of the building 
   2. Rohit and Neha 
   3. Neha and the tree 
   4. building and the tree.

Answer 1

a. From figure,

`tan 30^circ = (AB)/(AR)`

Substituting values, we get:

⇒ `1/sqrt(3) = (AB)/150`

⇒ `AB = 150/sqrt(3)`

⇒ AB = `150/1.732`

⇒ AB = 86.6 m

Hence, height of the building = 86.6 m.

b. 

i. From figure,

`tan 60^circ = (AB)/(AN)`

Substituting values, we get:

⇒ `sqrt(3) = (150/sqrt(3))/(AN)`

⇒ `AN = 150/sqrt(3) xx 1/sqrt(3)`

⇒ `AN = 150/3`

⇒ AN = 50 m

Hence, distance between Neha and foot of the building = 50 m.

ii. From figure,

RN = AR – AN

= 150 – 50

= 100 m

Hence, distance between Rohit and Neha = 100 m.

iii. From figure,

`tan 60^circ = (ST)/(NT)`

Substituting values, we get:

⇒ `sqrt(3) = 30/(NT)`

⇒ `NT = 30/sqrt(3)`

⇒ `NT = 30/sqrt(3) xx sqrt(3)/sqrt(3)`

⇒ `NT = (30sqrt(3))/3`

⇒ `NT = 10sqrt(3)`

⇒ NT = 10 × 1.732

⇒ NT = 17.32 m

Hence, distance between Neha and the tree = 17.32 m.

iv. From figure,

AT = AN – NT

= 50 – 17.32

= 32.68 m

Hence, distance between building and tree = 32.68 m.