At the foot of the mountain the angle of elevation of its summit is found to b $α$ . After ascending a ft. towards the mountain up a slope of inclination $β$ , the angle of elevation is found to be $γ$ . Show that the height of the mountain is
$\frac{a s i n α s i n (γ - β)}{s i n (γ - α)}$ feet.

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Solution

The angle of elevation of summit A at C is $α$ , and at D is $γ$ where CD = a and $∠ D C B = β$ .
From $Δ E A D, γ = ∠ D A E + α$ .
$∴ ∠ D A E = γ - α$
Also from $Δ C A D,$
$α - β + γ - α + ∠ C D A = 180^{o}$
$∴ ∠ C D A = 180^{o} - (γ - β)$
$∴ s i n C D A = s i n (γ - β)$
Now two sides h = AC and and a = CD are given.
They come in triangles ACB and ACD respectively and both have side AC common. Applying sine rule on both and eliminate common side AC
$\frac{h}{s i n α} = \frac{A C}{s i n 90^{o}} Δ A C B ∴ A C = \frac{h}{s i n α}$ ...(1)
$\frac{A C}{s i n C D A} = \frac{C D}{s i n C A D} Δ A C D$
or $A C = \frac{a s i n α s i n (γ - β)}{s i n (γ - α)}$ ...(2)
Hence from (1) by the help of (2).
$h = \frac{a s i n α s i n (γ - β)}{s i n (γ - α)}$

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