TF is a tower with F on the ground. The angle of elevation of T from A is x° such that tan x°= 2/5 and AF = 200 m. The angle of elevation of T from a nearer point B is y° with BF = 80 m. The value of y° is-

Answer: Option B

Solution

Answer: Option B

Solution:

$$\text{Given tan }{x}^{\circ }=\frac{2}{5}$$     and AF = 200 meter
$$\begin{array}{rl}& \therefore \frac{2}{5}=\frac{TF}{AF}\\ & \Rightarrow TF=\frac{2\times 200}{5}\\ & \Rightarrow TF=80\text{ m}\\ & \text{We have, BF = 80 m}\\ & \therefore \tan y^{\circ }=\frac{TF}{BF}\\ & \Rightarrow \tan y^{\circ }=\frac{80}{80}\\ & \Rightarrow \tan y^{\circ }=1=\tan 45^{\circ }\\ & \therefore y^{\circ }=45^{\circ }\end{array}$$