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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 - Diagram for the question 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.

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  • $\begingroup$ Whenever you type trig functions, use a backlash before the trig function. Example: \sin(x) instead of sin(x). "\sin(x)" will produce $\sin(x)$,i.e., the name of trig ratio is not italicized. :-) $\endgroup$ – Nirbhay Feb 21 '17 at 16:32
  • $\begingroup$ Okay. Shall take care of this the next time. Thanks. $\endgroup$ – Saksham Feb 21 '17 at 16:34
  • $\begingroup$ Have Fun!!! Good Luck.... ;-) $\endgroup$ – Nirbhay Feb 21 '17 at 16:35
  • $\begingroup$ You may also want to use block (four dollar signs) instead of the standard form. Normally you will write Math by enclosing it within two dollar signs, one at each end. Instead, try to enclose it within four dollar signs, two at each end. For Example : $\$\$<\text{Math}>\$\$$. You can see the edit by @lioness99a. He added blocks to your math. $\endgroup$ – Nirbhay Feb 21 '17 at 16:39
  • $\begingroup$ Thanks for the advice! Will keep it in mind the next time I post anything. $\endgroup$ – Saksham Feb 21 '17 at 17:25
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Cheers to you! Your answer is absolutely correct. Don't trust the book if you are sure of your answer!

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  • $\begingroup$ Alright. Thanks a lot for the boost! $\endgroup$ – Saksham Feb 21 '17 at 16:34
  • $\begingroup$ @Saksham (+1) to you for your correct solution! $\endgroup$ – Rohan Feb 21 '17 at 16:35

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