TF is a tower with F on the ground. The angle of elevation of T from A is x° such that tan x°= $$\frac{2}{5}$$   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-

Solution(By Examveda Team)

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