Let AB be the mountain, so the at the foot of a mountain the elevation of its summit is 45^o

So, ∠ACB = 45°

Now when moving on the slope of 30° by a distance of 1000m,

i.e., the CD is the distance moved on the slope of 30° towards the mountain,

Hence CD = 1000m………..(i)

And ∠DCF = 30°

Let EB = FD = x……..(ii)

DE = FB = z…….(iii)

CF = y and AE = t………(iv)

So after moving 1000m, the elevation becomes 60°,

So ∠ADE = 60°

In ΔDFC,

And

Hence

In ΔADE,

⇒t = z√3………(vi)

In ΔABC,

⇒t + x = y + z

⇒z√3 + 500 = 500√3 + z (from (iv), (v), (vi))

⇒z√3-z = 500√3-500

⇒z(√3-1) = 500(√3-1)

⇒z = 500m……(vii)

Hence the equation (vi) becomes,

t = z√3 = (500) √3 m

Hence the height of the mountain is

AB = AE + EB = t + x = (500√3 + 500)m

So the height of the mountain is 500(√3 + 1)m.