In the diagram, < PQR = < PSQ = 90°, |PS| = 9 cm,... | WAEC SSCE - General Mathematics - 1988 (Essay)

Question 1 Report

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|.

Answer

From the diagram, S lies on \(PR\) with \(\angle PQR = 90^\circ\) (right angle at Q) and \(\angle PSQ = 90^\circ\), so \(QS\) is the altitude from Q onto the hypotenuse \(PR\). The given lengths are \(|PS| = 9\text{ cm}\), \(|SR| = 16\text{ cm}\) and \(|SQ| = x\).

(a) Value of x by a trigonometric ratio

Let \(\angle QPS = \alpha\). In right triangle \(PSQ\):

\[\tan\alpha = \frac{QS}{PS} = \frac{x}{9}\]

In right triangle \(QSR\), the angle \(\angle SQR\) equals \(\alpha\) (both equal \(90^\circ - \angle R\)), so

\[\tan\alpha = \frac{SR}{QS} = \frac{16}{x}\]

Equating the two expressions:

\[\frac{x}{9} = \frac{16}{x} \;\Rightarrow\; x^2 = 9\times 16 = 144\]\[x = 12\text{ cm}\]

(b)(i) Size of \(\angle QRS\) to the nearest degree

In right triangle \(QSR\):

\[\tan(\angle QRS) = \frac{QS}{SR} = \frac{12}{16} = 0.75\]\[\angle QRS = \tan^{-1}(0.75) = 36.87^\circ \approx 37^\circ\]

(b)(ii) \(|PQ|\)

In right triangle \(PSQ\), by Pythagoras:

\[|PQ|^2 = |PS|^2 + |QS|^2 = 9^2 + 12^2 = 81 + 144 = 225\]\[|PQ| = \sqrt{225} = 15\text{ cm}\]

Therefore \(x = 12\text{ cm}\), \(\angle QRS \approx 37^\circ\) and \(|PQ| = 15\text{ cm}\).

Read lesson note on Introductory Calculus (WAEC)

Read lesson note on Trigonometry (JAMB)

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