Given:A pole of height 4 m is fixed on the top of a tower. The angle of elevation of the top of the pole as observed from a point on the ground is 60o.The angle of depression of the same point from the top of the tower is 45°.Formula used:For the right-angled triangle.tan(?) = \(\frac{p}{b}\)Calculation:According to the question, the required figure is:In ?ABC,tan(60?) = \(\frac{4+h}{b}\) ? ?3 = \(\frac{4+h}{b}\) ----(1)In ?DBC,tan(45?) = \(\frac{h}{b}\)? 1 = \(\frac{h}{b}\)? h = b ----(2)By using the equation (1) and (2), ? ?3 = \(\frac{4+h}{h}\)? ?3h = 4 + h? h × (?3 - 1) = 4? h = \(\frac{4}{?{3} -1} × \frac{?{3} +1}{?{3} +1}\)? h = \(\frac{4(?{3} +1)}{(?{3})^2 -1} \)? h = \(\frac{4(?{3} +1)}{(3 -1)} \)? h = 2 × (?3 + 1) m? 2(?3 + 1) is the required height (in m) of the tower.