(a) The angle of depression of a point P on the ground from the top T of a building is 23.6°. If the distance from P to the foot of the building is 50m, calculate, correct to the nearest metre, the height if the building.
(b)
In the diagram, (PT // SU, QS // TR, /SR/ = 6cm) and (/RU/ = 10 cm). If the area of (Delta TRU = 45 cm^{2}), calculate the area of the trapezium QTUS.
Explanation
(a)
(< P = 23.6°) (Alternate angle)
(tan 23.6 = frac{x}{50})
(x = 50tan 23.6)
= (50 times 0.4369)
= (21.845 m approxeq 22 m) (to the nearest metre)
(b) Given that (Delta TRU = 45cm^{2})
(implies frac{1}{2}(bh) = 45)
(10h = 45 times 2 = 90)
(h = 9cm)
Area of trapezium = (frac{1}{2}(a + b)h), where (a = 6cm; b = 10 + 6 = 16cm; h = 9cm)
= (frac{1}{2}(6 + 16) times 9)
= (11 times 9)
= (99 cm^{2})