RGS P2 Prelim Q6a 1997
A church tower stands on horizontal ground and a point T at the top of the tower is 25 metres vertically above a point M of the base. The points A and B are on the same level as M. The point A is due south of the tower and the angle of elevation of T from A is 27°. The point B is due west of the tower and the angle of elevation of T from B is 35°.
(i) Calculate to one decimal place, the distances of AM, BM and BA respectively.
(ii) Calculate to the nearest degree, the bearing of B from A.
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Answer:
(i)
25/AM = tan 27°
AM = 25/tan 27° = 49.07 m
25/BM = tan 35°
AM = 25/tan 35° = 35.7 m
BA2 = BM2 + AM2
Thus, BA = √(35.72 + 49.072)
BA = 60.7 m
(ii) angle MAB = tan-1 35.7 / 49.07 = 36.04°
Hence, bearing of B from A = 360° - 36.04° = 324°
Note: Angle BMA is 90° because B is west of M and A is south of M
