# A semicircular are AB of length 2 L and a vertical tower PQ are situated in the same vertical plane. The feet A and B of the arch and the base Q of the tower are at the same horizontal level, with B between A and Q. A man at A finds the tower hidden from his view due to the arch. He starts crawling up the arch and just sees the topmost point P of the tower after covering a distance L/2 along the arch. He crawls further to the topmost point of the arch and notes the angle of elevation of P to be θ. Compute the height of the tower in terms L and θ.

Jitender Pal
8 years ago
Hello Student,
Let AMNB be the semicircular arc of length 2L, and PQ be the vertical tower so that A, B, Q are in the same horizontal line.
Let M be the pt. on arc s. t. AB = L/2 then as AM = 1/4 AB we should have ∠AOM = 45°, As at M the man just sees the top most pt P of the tower, tangent through M must pass through P and hence ∠OMP = 90° and then by simple geometry we get ∠PMY = 45°.
Also N is the top most pt. of arc AB, hence ON must be vertical.
∴ ON = r = XQ
∴ PX = h – r, where h is ht of tower PQ, and OK = YQ = OM cos 45° = r/√2
Similarly, MK = OM sin 45° = r√2
∴ In ∆ PMY we get tan 45° = PY/MY
⇒ PY = MY
⇒ h – QY = MK + KY
⇒ h – OK = r/√2 + x
⇒ h – r/√2 + r/√2 + x
⇒ x = h – r √2 . . . . . . . . . . . . . . . . . . . . . . . . . (1)
In ∆ PNX, tan θ = PX/NX
⇒ tan θ = (h – r)/x
⇒ x = (h – r) cot θ . . . . . . . . . . . . . . . . (2)
Comparing the values of x from (1)and (2), we get
h – r √2 = h cot θ – r cot θ
h = r (√2 – cot θ)/(1 – cot θ)
But are length = 2L = π r ⇒ r = 2L/π
∴ h = 2L/π [√2 – cot θ/1 – cot θ]

Thanks
Jitender Pal