Let P be the top and M be the foot of the mountain PM.
Let the observer be at the observer ascends along the mountain at slope AB such that `angleMAB=30^(@)` and AB = 1000 m. Therefore, B is the second position of the observer, where the angle of elevation `angleNBP=60^(@)`.
We have, `angleMAP=angleMPA=45^(@)`,
`angleMAB=30^(@),angleNPB=30^(@)`
Therefore, `angleBAP=angle=15^(@)`
Hence, the triangleABP is iosceles and BP = AB = 1000 m (sides opposite to equal angles are equal)
Now, `PM-PN+NM=PN+BC`
`=BP sin 60^(@)+ABsin30^(@)`
`=1000(sqrt3/2)+1000(1/2)=500(sqrt3+1)`
Hence, the height of the tower is `500(sqrt3+1)`m.
![image](https://learnqa.s3.ap-south-1.amazonaws.com/images/1611323910143005034fu1iaZMy4MGHvhg4.png)