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.
A. 75° B. 45° C. 60° D. 30° Answer: Option BShow Answer
Solution(By Apex Team)
$\begin{array}{l}\text{Given}\\
\text{tan}\ x^{\circ}=\frac{2}{5}\\
\text{and AF}=200\ \text{metre}\\
\therefore \frac{2}{5}=\frac{T F}{A F} \\
\Rightarrow T F=\frac{2 \times 200}{5} \\
\Rightarrow T F=80 \mathrm{~m} \\
\text { We have, } \mathrm{BF}=80 \mathrm{~m} \\
\therefore \tan y^{\circ}=\frac{T F}{B F} \\
\Rightarrow \tan y^{\circ}=\frac{80}{80} \\
\Rightarrow \tan y^{\circ}=1=\tan 45^{\circ} \\
\therefore y^{\circ}=45^{\circ}
\end{array}$
HI 
EN 