From the top of a building 20 m high, the angle of elevation of the top of a tower is 30° and the angle of depression of its foot is 45°. Find the distance between the building and the tower, and the height of the tower. (\(\sqrt3 = 1.73\))

Answer: Distance = 20 m; height of tower \(= 20 + \dfrac{20}{\sqrt3} \approx 31.53\) m

[Figure: Building AB = 20 m (top A, foot B); tower CD (top D, foot C). From A draw AE horizontal, meeting CD at E. \(\angle DAE = 30^\circ\), \(\angle EAC = 45^\circ\).]

ABCE is a rectangle, so EC = AB = 20 m and AE = BC.

In right-angled ΔAEC: \(\tan 45^\circ = \dfrac{EC}{AE}\) \(\therefore 1 = \dfrac{20}{AE}\) \(\therefore AE = 20\) m = BC.

In right-angled ΔAED: \(\tan 30^\circ = \dfrac{DE}{AE}\) \(\therefore DE = \dfrac{20}{\sqrt3} = \dfrac{20\sqrt3}{3} = \dfrac{34.6}{3} \approx 11.53\) m.

Height of tower CD = CE + ED = 20 + 11.53 = 31.53 m.

\(\therefore\) The building and tower are 20 m apart, and the tower is about 31.53 m high.