Consider the following for the two (02) items that follow: The top ( M ) of a tower is observed from three points P , Q and R lying in a horizontal straight line which passes directly along the foot ( N ) of the tower. The angles of elevations of M from P , Q and R are 30° , 45° and 60° respectively. Let PQ = a and QR = b . What is PN equal to?

Step-by-step solution

Let h be the height of the tower MN . In MNP , 30^ = h PN PN = h 3 In MNQ , 45^ = h QN QN = h In MNR , 60^ = h RN RN = h 3 Since the angles of elevation increase as we move closer to the tower, the points P, Q, R lie on the same side of N in the order P, Q, R, N . Given PQ = a , we have PN - QN = a . h 3 - h = a h( 3 - 1) = a h = a 3 - 1 Rationalizing the denominator, we get: h = a( 3 + 1) ( 3 - 1)( 3 + 1) = a( 3 + 1) 2 We need to find PN , which is h 3 : PN = a( 3 + 1) 2 3 = ( 3 + 3 2 )a