A vertical pole subtends an angle $ an ^{-1}\left(\frac{1}{2} ight)$ at a point $P$ on the ground. If the angles substended by the upper half and the lower half of the pole at $P$ are respectively $\alpha$ and $\beta$, then $( an \alpha, an \beta)$ is equal to

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A vertical pole subtends an angle $	an ^{-1}\left(\frac{1}{2}ight)$ at a point $P$ on the ground. If the angles substended by the upper half and the lower half of the pole at $P$ are respectively $\alpha$ and $\beta$, then $(	an \alpha, 	an \beta)$ is equal to

Aicharya · Accepted Answer

Let $A C$ be a pole and point $P$ be the position on of the ground. Given, $	heta=	an ^{-1} \frac{1}{2}$ $\Rightarrow \quad 	an 	heta=\frac{1}{2}$ Also, $\quad 	heta=\alpha+\beta$ $\begin{aligned} \Rightarrow & 	an 	heta & =	an (\alpha+\beta) \ \Rightarrow & \frac{1}{2} & =\frac{	an \alpha+	an \beta}{1-	an \alpha 	an \beta}\end{aligned}$ (a) When $(	an \alpha, 	an \beta)=\left(\frac{1}{4}, \frac{1}{5}ight)$ $	herefore$ RHS $=\frac{\frac{1}{4}+\frac{1}{5}}{1-\frac{1}{4} 	imes \frac{1}{5}}=\frac{\frac{9}{20}}{\frac{19}{20}}$ $=\frac{9}{19} 
eq \frac{1}{2}$, not true (b) $\begin{aligned} & 	ext { When }(	an \alpha, 	an \beta)=\left(\frac{1}{5}, \frac{2}{9}ight) \ & \begin{aligned} 	ext { RHS } & =\frac{\frac{1}{5}+\frac{2}{9}}{1-\frac{1}{5} 	imes \frac{2}{9}}=\frac{\frac{19}{45}}{\frac{43}{45}} \ = & \frac{19}{43} 
eq \frac{1}{2}, 	ext { not true }\end{aligned}\end{aligned}$ (c) When $(	an \alpha, 	an \beta)=\left(\frac{2}{9}, \frac{1}{4}ight)$ RHS $=\frac{\frac{2}{9}+\frac{1}{4}}{1-\frac{2}{9} 	imes \frac{1}{4}}=\frac{\frac{17}{36}}{\frac{34}{36}}=\frac{1}{2}$, true

A vertical pole subtends an angle $\tan ^{-1}\left(\frac{1}{2}\right)$ at a point $P$ on the ground. If the…

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