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Solution :

Let AB be a tower of height h. C and D be two points at distance 4 m and 9 m respectively from B. <br> Let `angleADC=theta` <br> `:. angleACB=90^(@)-theta` <br> In `DeltaABD` <br> `tantheta=(AB)/(BD)=h/9 …(1)` <br> In `Delta ABC` <br> `tan(90^(@)-theta)=(AB)/(BC)rArr cottheta=h/4` <br> `rArr 1/tantheta=h/4 rArr 9/h=h/4" [from(1)]"` <br> `rArr h^2-36 rArr h-6` <br> `:. " Height of the tower "=6 m` <br> OR <br> In the above question, put `theta=30^(@)` <br> So that `90^(@)-theta=60^(@)` and hence on change in the solution. <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NTN_MATH_X_C09_S01_039_S01.png" width="80%">