The sum of angles of elevation of the top of a tower from two points distant $a$ and $b$ from the base and in the same straight line with it is $90^{\circ}$. Then, the height of the tower is

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Aicharya · Accepted Answer

Given, $	heta+\phi=90^{\circ}$ Let the height of a tower $=h$ Then, In $	riangle A C P$ $	an 	heta=\frac{h}{a}$ ...(i) In $	riangle A B P$ $	an \phi=\frac{h}{b}$ ...(ii) $\because \quad 	an (	heta+\phi)=\frac{	an 	heta+	an \phi}{1-	an 	heta \cdot 	an \phi}$ $	an 90^{\circ}=\frac{\frac{h}{a}+\frac{h}{b}}{1-\frac{h}{a} \cdot \frac{h}{b}}=\infty$ $\Rightarrow \quad 1-\frac{h^2}{a b}=0$ $\Rightarrow \quad h^2=a b$ $\Rightarrow \quad h=\sqrt{a b}$ Hence, the required height of a tower is $\sqrt{a b}$.

The sum of angles of elevation of the top of a tower from two points distant $a$ and $b$ from the base and…

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