Hello again arze Originally Posted by

**arze** 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

I called the foot of the mast $\displaystyle P$, its top $\displaystyle Q$, the point immediately opposite to it $\displaystyle R$, and the mid-point of $\displaystyle AB,\; S$. The height of the mast I denoted by $\displaystyle h$. So, from part (i):$\displaystyle h = \frac{4a}{\sqrt{\cot^2\beta-\cot^2\alpha}}$

$\displaystyle \Rightarrow \frac{a^2}{h^2}= \frac{\cot^2\beta - \cot^2\alpha}{16} $ ...(1)

Then $\displaystyle \angle PSQ = \theta$ and, from $\displaystyle \triangle PQS$:$\displaystyle PS = h\cot\theta$

and from $\displaystyle \triangle PSR$:$\displaystyle PS^2 = PR^2+SR^2$

$\displaystyle \Rightarrow h^2\cot^2\theta = h^2\cot^2\beta - 25a^2 + a^2$, using the result from (i) that $\displaystyle PR^2 =h^2\cot^2\beta - 25a^2$

$\displaystyle \Rightarrow \cot^2\theta = \cot^2\beta - \frac{24a^2}{h^2}$$\displaystyle =\cot^2\beta - \frac{3(\cot^2\beta-\cot^2\alpha)}{2}$, from (1)

$\displaystyle \Rightarrow 2\cot^2\theta = 3\cot^2\alpha - \cot^2\beta$

Grandad