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|.
(a) Find the value of x using a trigonometric ratio.
(b) Calculate : (i) the size of < QRS to the nearest degree; (ii) |PQ|.
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Correct Answer: Option n
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,
\(|PQ|^{2} = 12^{2} + 9^{2}\)
= \(144 + 81 = 225\)
\(|PQ| = \sqrt{225} = 15 cm\)
(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,
\(|PQ|^{2} = 12^{2} + 9^{2}\)
= \(144 + 81 = 225\)
\(|PQ| = \sqrt{225} = 15 cm\)