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A and B are 2 points on level ground, and B is a metres due east of A; a tower, h metres high , is also on the same level ground. From A the tower is in a direction of N θ E and from B it is N $\phi$ W. From the top of the tower, the angle of depression of A is α and of B it is β. Prove:

$(i) h\sin(\theta + \phi) = a\cos\phi\ \tan\alpha \\(ii) cos\phi \tan\alpha = cos\theta\tan\beta \\(iii)h^2(\cot^2\alpha - \cot^2\beta) - 2ha\cot\alpha\sin\theta + a^2 = 0$

I think the question may be wrong. I do not find either (i) or (ii) true so haven’t proceeded to (iii). I should be grateful if anyone could confirm or refute i or ii.

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  • $\begingroup$ There are two small errors in your sketch relating to the angles $\alpha$ and $\beta$. The angle of depression is measured relative to the horizontal, so in fact the angles that you have marked $\alpha$ and $\beta$ should be replaced with $90^{\circ} - \alpha$ and $90^{\circ} - \beta$ respectively. $\endgroup$ – Zac Nov 19 '19 at 13:22
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With the corrections from my comment in mind, the first result can be shown by considering $\triangle APT$ and noting that $$\tan{\alpha} = \frac{h}{AP}.$$ Applying the sine rule in $\triangle APB$ (noting that $\angle APB = \theta + \phi$) gives $$\frac{a}{\sin{(\theta + \phi)}} = \frac{AP}{\cos{\phi}}$$ as you have noted, and combining these should give the required result.


The second result should follow by symmetry, since the same argument from part (i) shows that $h \sin{(\theta + \phi)} = a \cos{\theta} \tan{\beta}.$

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