Let AB be the peak of a hill, so the at the station A the elevation of its summit is α°

So, ∠CAB = α°

Now when moving on the slope of β° by a distance of ‘c’m,

i.e., AD is the distance moved on the slope of β° towards the hill,

Hence AD = ’c’m………..(i)

And ∠DAF = β°

Let EB = FD = x……..(ii)

DE = FB = z…….(iii)

AF = y and CE = t………(iv)

So after moving ‘c’m, the elevation becomes γ°,

So ∠CDE = γ°

In ΔDFA,

And

In ΔCDE,

⇒z = t cot γ………(vi)

In ΔCBA,

⇒tan α (c cos β + t cot γ ) = t + c sin β

⇒(c tan α cos β + t tan α cot γ ) = t + c sin β

⇒t-t tan α cot γ = c tan α cos β-c sin β

⇒t(1 - tan α cot γ ) = c(tan α cos β-sin β )

Now,

AB = AE + EB = t + x

So the height of the hill is

Hence proved