Doubt by Zoha
Solution :
AB=2BP (Given)
AB=50 m (Given)
2BP=50 m
BP=50/2
BP=25 m
In Δ ABC
∠1+θ = 2θ [Exterior angle property of a triangle]
∠1 = 2θ-θ
∠1 = θ
∠1 = θ
Now
∠BAC=∠BCA = θ
AB = BC (Sides opposite to equal amgles of a triangle are equal)
∠BAC=∠BCA = θ
AB = BC (Sides opposite to equal amgles of a triangle are equal)
In Rt. ΔBPC
cos2θ=B/H
cos2θ=BP/BC
cos2θ=BP/AB [∵ BC=AB]
cos2θ=BP/2BP [∵ AB=2BP]
cos2θ=1/2
cos2θ=cos60°
2θ = 60°
cos2θ=BP/BC
cos2θ=BP/AB [∵ BC=AB]
cos2θ=BP/2BP [∵ AB=2BP]
cos2θ=1/2
cos2θ=cos60°
2θ = 60°
Hence, the angle of elevation of the top of the tower from B is 60°
Again, In Rt. Δ BPC
tan2θ = P/B
tan2θ =PC/BP
tan60°=PC/25
√3 = PC/25
PC = 25√3
PC = 25×1.732
PC = 43.3 m
Hence, the height of the tower is 43.3 m
tan2θ =PC/BP
tan60°=PC/25
√3 = PC/25
PC = 25√3
PC = 25×1.732
PC = 43.3 m
Hence, the height of the tower is 43.3 m
Note: This can also be solved by using the formula, tan2θ=2tanθ/(1-tan²θ) but this formula is belongs to class XI Mathematics.
In Rt. ΔBPC
tan2θ=PC/BP
tan2θ=h/25
h=25tan — (1)
In Rt. ΔAPC
tanθ=PC/AP
tanθ=h/(AB+BP)
tanθ=h/(50+25)
tanθ=h/75
tanθ=25tan2θ/75 [from eq (1)]
tanθ=tan2θ/3
3tanθ=tan2θ
3tanθ=2tanθ/(1-tan²θ)
3=2/(1-tan²θ)
3×(1-tan²θ)=2
3-3tan²θ=2
3-2=3tan²θ
1=3tan²θ
1/3=tan²θ
1/√3=tanθ
tan30°=tanθ
30°=θ
θ=30°