Question

A chimney leans towards North. At equal distances due North and South of it in a horizontal plane, the elevations of the top are $\alpha ,\beta$. Show that the inclination of the chimney to the vertical is
${\tan }^{-1}\left[\frac{\sin (\alpha -\beta )}{2\sin \alpha \sin \beta }\right]$.

Answer 1

Applying $(m-n)$ theorem of trigonometry on $△PQA$ in which angle

$\angle PQA={90}^o-\theta$

$(a+a)\cot ({90}^o-\theta )=a\cot \beta -a\cot \alpha$

$\therefore 2\tan \theta =\frac{\sin \left(\alpha -\beta \right)}{\sin \alpha \sin \beta }$

$\therefore \theta ={\tan }^{-1}\left(\frac{\sin \left(\alpha -\beta \right)}{\sin \alpha \sin \beta }\right)$

From $\left(1\right)$ we can also say that

$\theta ={\tan }^{-1}\frac{1}{2}(\cot \beta -\cot \alpha )$