# (i) PQ is a vertical tower. P is the foot and Q is the top of the tower. A, B, C are three points in the horizontal plane through P. The angles of elevation of Q from A, B, C are equal, and each is equal to θ. The sides of the triangle ABC are a, b, c  and the area of the triangle ABC is ∆. Show that the height of the tower isabc tan θ/A∆(ii) AB is a vertical pole. The end A is on the level ground. C is the middle point of AB. P is a point on the level ground. The portion CB subtends an angle β at P. If AP = n AB, then show that tan β = n/2n2 + 1

8 years ago
Hello Student,
(i) Let h be the height of tower PQ.
In ∆ APQ tan θ = h/AP ⇒ AP = h/tan θ
Similarly in ∆’s BPQ and CPQ we obtain
BP = h/tan θ = CP
∴ AP = BP = CP
⇒ P is the circum centre of ∆ ABC with circum radius R = AP = abc/4∆
∴ h = AP tan θ = abc tan θ/4 ∆
(ii) Given AP = AB x n
⇒ AB/AP = 1/n = tan θ
∴ tan θ = 1/n
Also tan (θ – β) = AC/AP = 1/2 AB/AP = 1/2n
⇒ tan θ – tan β/ 1+ tan θ tan β = 1/2n ⇒ $\frac{\frac{1}{n}-tan\beta }{1+\frac{1}{n}tan \beta}$= 1/2n
⇒ 2n – 2n2 tan β = n + tan β
⇒ (2n2 + 1) tan β = n ⇒ tan β = n/2n2 + 1

Thanks