Question

A right circular cylinderical tower of height h and radius r stands on a horizontal plane. Let A be the point in the horizontal plane and PQR be the semicircular edge of the top of the tower such that Q is the point in it nearest to A. The angles of elevation of the points P, R. and Q from A are 45°, 45° and 60° respectively. Show that

h/r = √3(1 + √5)/2

Answer 1

Let P', Q', R' be the points on the base of the cylinder corresponding to the points, P, Q, R on the top. OO' is the axis of the cylinder.

. ^.. PP' = QQ' = OO' = RR' = h

Since, the point Q in the semicircular edge PQR is nearest to A, therefore Q' is on the line AO'.

Since, AP'= AR' and O' is the mid point of P' R'.

So, from the right angled ΔAO' R'.