I saw a question in a book and sat down to solve it. It is based on the applications of trigonometry. I solved but the answer in the book differs slightly than mine. Here's the question and my subsequent attempt:
A chimney leans towards the north and its angles of elevation when observed from equal distance from north and south of the chimney in the same horizontal plane are found to be $ \alpha $ and $ \beta $ respectively. If the inclination of the chimney to the vertical be $\theta$, then $\tan\theta$ is?
Here's my diagram -
My attempt-
$BC = h\tan\theta$ \begin{align} \tan\alpha &=\frac {h}{x-h \tan\theta}\\ x \tan\alpha - h\tan\theta\tan\alpha &= h\\ x \tan\alpha &= h (1 + \tan\theta\tan\alpha)\\ x &= h\cdot\frac{1+ \tan\theta\tan\alpha}{\tan\alpha}\\ &= h (\cot\alpha + \tan\theta)\tag{1}\label{1} \end{align}
Similarly, \begin{align} \tan\beta &= \frac{h}{x+h\tan\theta}\\ x \tan\beta + h\tan\theta\tan\beta &= h\\ x \tan\beta &= h(1-\tan\theta\tan\beta)\\ x &= h\cdot\frac{1-\tan\theta\tan\beta}{\tan\beta}\\ x &= h (\cot\beta - \tan\theta)\tag{2}\label{2} \end{align}
Now, from \eqref{1} and \eqref{2}: \begin{align} h (\cot\beta - \tan\theta) &= h (\cot\alpha + \tan\theta)\\ \cot\beta - \cot\alpha &= 2 \tan\theta\\ \tan\theta &= \frac{\cot\beta - \cot\alpha}{2}\\ &= \frac{1}{2}\left[\frac{\cos\beta}{\sin\beta}-\frac{\cos\alpha}{\sin\alpha}\right]\\ \tan\theta &= \frac {\sin\alpha\cos\beta - \cos\alpha\sin\beta}{2\sin\beta\sin\alpha}\\ \tan\theta &= \frac {\sin(\alpha - \beta)} {2\sin\alpha\sin\beta} \end{align}
But, the answer given in the book is $$\tan\theta = \frac {\sin(\alpha + \beta)} {2\sin\alpha\sin\beta}\ .$$
Have I gone wrong anywhere in my solution? Or is the book's answer wrongly printed? If I have gone wrong anywhere, please let me know. I'll be grateful. If the book's answer is correct, then please provide a solution as to how we'll get that answer.
Thanks.
