ABC is a triangular park with AB = AC = 100 m. A block tower is situated at the midpoint of BC.The angles of elevation of the top of the tower at A and B are `cot^-1(3.2)` and `cosec^-1(2.6)` respectively.The height of the tower is:

Explanation

In the figure DP is a clock tower of height 'h' standing at D, the midpoint of BC.

since AB=AC ,AD is perpendicular to BC. In triangle ADP.
`tan alpha = (h)/(AD)`
`rArr AD cot alpha`
In triangle PDB ,
`tan beta =(h)/(BD)`
`rArr BD = h cot beta `
Now ,in right triangle ADB,
` AB^2+AD^2+BD^2`
`rArr " " 100^2=h^2 cot ^2 alpha + h^2 cot ^2+ beta`
`rArr " " 100^2= h^2 [cot^2 alpha + (cosec ^2 beta -1)]`
`rArr " " 100^2=h^2[(3.2)^2+(2.6)^2-1)]`
`rArr " " 100^2=16 h^2`
`rArr " " h^2=625`
`rArr " " h=25 m`