Question

The angle of elevation of the top a cliff from a fixed point A is θ. After going up a distance of K meters towards the top of the cliff at an angle of ϕ. It is found that the angle of elevation is α. show that the height of the cliff in meters is K (kcosϕ - sinϕcosα)/(cotθ - cosα).

Answer 1

Let CD is a hill. An angle of elevation at point A and B are ϕ, θ and α respectively.

AB = K m

∠BAM = ϕ

∠DAC = θ and ∠DBN = α

Let CD = h m

From right angled ∆AMB

cos ϕ = AM/AB

⇒ AM = AB cos ϕ = K cos ϕ …(i)

and sin ϕ = BM/AB

⇒ BM = AB sin ϕ = K sin ϕ …(ii)

From right angled ∆ACD,

cos θ = AC/DC

⇒ AC = DC cot θ = h cot θ …(iii)

∵ MC = AC – AM

From eq. (i) and (iii),

MC = h cot θ – K cos ϕ

∵ MC = BN

So, BN = h cot θ – K cos ϕ …..(iv)

From right angled ∆BND,

tan α = DN/BN

⇒ DN = BN tan α

= MC tan α

= (h cot θ – K cos ϕ) tan α

h = DN + NC

= h cot θ tan α – K cos ϕ tan α + K sin ϕ

h[1 – cot θ tan α] = K[sin ϕ – cos ϕ tan α]