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:

(a) height of the building

(b) distance between

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

(a) From figure,

tan 30° = $\dfrac{AB}{AR}$

Substituting values we get :

$\Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AB}{150} \\[1em] \Rightarrow AB = \dfrac{150}{\sqrt{3}} = \dfrac{150}{1.732} \\[1em] = 86.6 \text{ m}.$

Hence, height of the building = 86.6 m.

(b)

1. From figure,

tan 60° = $\dfrac{AB}{AN}$

Substituting values we get :

$\Rightarrow \sqrt{3} = \dfrac{\dfrac{150}{\sqrt{3}}}{AN} \\[1em] \Rightarrow AN = \dfrac{150}{\sqrt{3}} \times \dfrac{1}{\sqrt{3}} \\[1em] \Rightarrow AN = \dfrac{150}{3} = 50\text{ m}.$

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

2. From figure,

RN = AR - AN = 150 - 50 = 100 m.

Hence, distance between Rohit and Neha = 100 m.

3. From figure,

tan 60° = $\dfrac{ST}{NT}$

Substituting values we get :

$\Rightarrow \sqrt{3} = \dfrac{30}{NT} \\[1em] \Rightarrow NT = \dfrac{30}{\sqrt{3}} \\[1em] \Rightarrow NT = \dfrac{30}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow NT = \dfrac{30\sqrt{3}}{3} \\[1em] \Rightarrow NT = 10\sqrt{3} \\[1em] \Rightarrow NT = 10 \times 1.732 = 17.32 \text{ m}.$

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.