Engineering
Mathematics
problems on height and distance
Question

PQR is a triangular park with PQ = PR = 200m. A T.V. tower stands at the mid point of QR. If the angles of elevation of the top of the tower at P, Q and R are respectively 45°, 30° and 30°, then the height of the tower (in m) is:

100

50

1003

502

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Solution

Let height of tower be h

∴ QM = h cot 30° =3h

and PM = h cot 45° = h

∴ PQ2 = PM2 + QM2

(200)2 = 3h2 + h2

h = 100 m.