The angle of elevation on the top of a mountain from a point on the ground is found to be $\alpha$. After walking a distance of $a$ along a slope of inclination $\beta$ towards the cliff, the angle of elevation is found to be $\gamma$. Show that the height of the mountain is
$$\frac{a \sin \alpha \sin (\alpha-\beta)}{\sin(\gamma-\alpha)}$$ .
I tried it this way:
From here, I calculated $BE$ to be
$$\frac{a\sin(\alpha-\beta)}{sin(\gamma-\alpha)}$$
Also, I got $FE$ to be $a\sin(\alpha-\beta)$.
Height $AO$ is $a\sin\alpha$ from the image.
I am unable to proceed hereafter. Any help would be appreciated. Thanks in advance.
Best Answer
The formula for the height is wrong.
We know $\angle{BCD}=\frac{\pi}{2}-\gamma$ and $\angle{BCA}=\frac{\pi}{2}-\alpha$ so therefore $\angle{DCA}=\frac{\pi}{2}-\alpha - (\frac{\pi}{2}-\gamma)=\gamma-\alpha$. We also know $\angle{DAC}=\alpha-\beta$. We therefore know that $\angle {ADC}=\pi +\beta-\gamma$. Using the law of sines we can find that $$\frac{\sin (\angle{ADC})}{AC} = \frac{\sin(\angle{DCA})}{a}$$ or $$\frac{\sin (\pi+\beta-\gamma)}{AC} = \frac{\sin(\gamma-\alpha)}{a}$$ or $$AC= \frac{a\sin(\gamma-\beta)}{\sin(\gamma-\alpha)}$$
As the height of the mountain is $h=AC\sin(\alpha)$, we arrive at the formula $$h =\frac{a\sin(\alpha)\sin(\gamma-\beta)}{\sin(\gamma-\alpha)}$$