Question

At a point on a level plane, a vertical tower subtends and angle $\alpha$ and a pole of height h meters at the top of the tower subtends an angle $\beta$. Show that the height of the tower is h $\frac{sin\beta .cos\alpha }{cos(\alpha +\beta )}$

Answer 1

In $\Delta ACP,tan\alpha =\frac{h}{y}\Rightarrow y=\frac{h}{tan\alpha }....\left(1\right)$
In $\Delta ADP,tan(\alpha +\beta )=\frac{h+x}{y}\Rightarrow y=\frac{h+x}{tan(\alpha +\beta )}....\left(2\right)$
From (1) & (2)
$\frac{h}{tan\alpha }=\frac{h+x}{tan(\alpha +\beta )}$
$\Rightarrow \frac{h+x}{h}=\frac{tan(\alpha +\beta )}{tan\alpha }\Rightarrow \frac{x}{h}=\frac{sin(\alpha +\beta )}{cos(\alpha +\beta )}.\frac{cos\alpha }{sin\alpha }-1$
$\frac{x}{h}=\frac{sin(\alpha +\beta ).cos\alpha -cos(\alpha +\beta ).sin\alpha }{cos(\alpha +\beta ).sin\alpha }\Rightarrow \frac{x}{h}=\frac{sin(\alpha +\beta -\alpha )}{cos(\alpha +\beta ).sin\alpha }\Rightarrow =h\frac{sin\beta .cos\alpha }{cos(\alpha +\beta )}$