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Bearing and angle between planes

Two questions i have partly done.

1. Prove that in a plane triangle ABC

$\displaystyle \cot A=\frac{a}{b}\csc C-\cot C$

The bearing of an aeroplane from a fixed observation post is $\displaystyle 052^o$, the aeroplane being at a horizontal distance of 6 miles from the post. Calculate the flight direction of the aeroplane if, when it passes due east of the post, it is at a horizontal distance of 9 miles from the post. If the height of the aeroplane when its bearing is $\displaystyle 052^o$ is *h* and its angle of elevation from the post is the same in both positions calculate, in terms of *h*, the height of the aeroplane when due east of the post.

In the figure

I first fin the length AB=$\displaystyle \sqrt{6^2+9^2-2\times 6\times 9\times\cos 38^o}=5.65$

then angle PAX=[math\frac{9\sin 38^o}{5.65}=78.9^o[/tex]

Then using the opposite angles on parallel lines, the direction of the plane would be $\displaystyle 180^o-(78.9^o-52^o)=153.1^o$

Answer is supposed to be 131

2. A vertical mast stands on the north bank of a river with straight parallel banks running east-west. The angle of elevation of the top of the mast is $\displaystyle \alpha$ when measured from A and point on the south bank *3a* to the east of the mast and $\displaystyle \beta$ when measured from another point B on the south bank distant *5a* to the west of the mast. Prove that the height of the mast is $\displaystyle \frac{4a}{\sqrt{\cot^2\beta-\cot^2\alpha}}$ and that the angle of elevation $\displaystyle \theta$ measured from a point midway between A and B is given by the equation $\displaystyle 2\cot^2\theta=3\cot^2\alpha-\cot^2\beta$

I have proved $\displaystyle \frac{4a}{\sqrt{\cot^2\beta-\cot^2\alpha}}$.

But i don't know how to begin with the next part

Thanks