Question

At the foot of a mountain the elevation of its summit is 45°.After ascending 1000m towards the top of the mountain at an angle of 30°,the elevation is found to be 60°.Find the height of the mountain.

Answer 1

Let point A be the position of summit of the mountain and B being its foot.Let C be the original position of observer and D, the final position after ascending 1000 metres.Let DN and DM be perpendiculars to BC and AB respectively.Thus,
CD = 1000 m 
∠DCN = 30⁰ 
and, ∠ADM = 60⁰
∠ACB = 45⁰ = ∠ACB 
∠DAM = 30° 
⇒ ∠DCA = 15° , ∠DAC = 15°
⇒ ∠DCA = ∠DAC 
⇒AD=CD = 1000 m 
now, 
In right Δ DCN 
sin 30° = DN/CD 
⇒1/2 = DN/1000
⇒ DN = 500 m 
And, In right Δ ADM 
sin 60° = AM/AD 
⇒ √3/2 = AM/1000 
⇒AM = 500 √3 = 500 × 1.732 = 866
⇒AM = 866 m 
TOTAL HEIGHT (AB) = BM + AM = DN + AM = 500 + 866 = 1366 m = 1.36.6 KM 

INDEX = / = UPON MEANS DIVIDE