Hello again cloud5 Originally Posted by

**cloud5** ...2. A man observed the top of a tower from the points A, B and C along a straight, horizontal road, where AB=a and BC=b. The angle of elevation of the top of the tower from A is $\displaystyle \alpha$ while the angle of elevation from B and C are both equal to $\displaystyle \beta$. If $\displaystyle \vartheta$ is the angle of elevation of the top of the tower from the point E wich is the mid-point of BC, show that

i)$\displaystyle \vartheta$ is the largest angle of elevation,

ii) $\displaystyle 4a(a+b)cot^2 \vartheta = (2a+b)^2 cot^2 \beta - b^2 cot^2 \alpha$

... and here's the outline for number 2.

Suppose that $\displaystyle P$ is the foot and $\displaystyle T$ the top of the tower, and that $\displaystyle TP = h$. Then we have: $\displaystyle \angle TAP = \alpha$

$\displaystyle \angle TBP = \angle TCP = \beta$

$\displaystyle \angle TEP = \theta$

$\displaystyle AB = a, BE = EC = \tfrac12b$

For part (i): $\displaystyle PB = h\cot\beta = PC$$\displaystyle \Rightarrow \triangle BPC$ is isosceles

$\displaystyle \Rightarrow PE \perp BC$

$\displaystyle \Rightarrow PE$ is the shortest distance from the foot of the tower to $\displaystyle BC$.

So?

(ii) $\displaystyle PE^2 = h^2\cot^2\theta = PB^2-BE^2$$\displaystyle \Rightarrow ...$

$\displaystyle \Rightarrow h^2 = \frac{b^2}{4(\tan^2\beta-\cot^2\theta)}$ (1)

$\displaystyle AP^2 = h^2\cot^2\alpha = PE^2+AE^2$$\displaystyle \Rightarrow ...$

$\displaystyle \Rightarrow h^2 = \frac{(2a+b)^2}{4(\cot^2\alpha - \cot^2\theta)}$ (2)

Eliminate $\displaystyle h$ from (1) and (2) and the result follows.

Can you fill in the gaps?

Grandad