![](https://www.sarthaks.com/?qa=blob&qa_blobid=14595922129126947172)
Let AB be the mountain, so the at the foot of a mountain the elevation of its summit is 45°
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,
![](https://www.sarthaks.com/?qa=blob&qa_blobid=6802497363903455403)
Hence y = 500√3 m.....(v)
In ΔADE,
![](https://www.sarthaks.com/?qa=blob&qa_blobid=9543047931849885791)
⇒ 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.