In the diagram, < PQR = < PSQ = 90°, |PS| = 9 cm, |SR| = 16 cm and |SQ| = x cm.
(a) Find the value of x using a trigonometric ratio.
(b) Calculate : (i) the size of < QRS to the nearest degree; (ii) |PQ|.
Explanation
(a)
Let (< SQR = theta ; < QRS = 90° – theta ; < QPS = theta)
In (Delta) PQS, (tan theta = frac{x}{9} … (1))
In (Delta) QRS, (tan theta = frac{16}{x} … (2))
From (1) and (2), (frac{x}{9} = frac{16}{x})
(x^{2} = 16 times 9 implies x = sqrt{16 times 9} = 12 cm)
(b) (i) From (Delta) QRS,
(tan theta = frac{16}{12} = 1.333)
(theta = tan^{-1} (1.333) = 53.1°)
(therefore < QRS = 90° – 53.1° = 36.9°)
(approxeq 37°) (to the nearest degree)
(ii) In (Delta) PQS,
(|PQ|^{2} = 12^{2} + 9^{2})
= (144 + 81 = 225)
(|PQ| = sqrt{225} = 15 cm)