The diagram shows triangle \(PQR\) with angle \(PQR = 90^\circ\), angle \(QPR = 37^\circ\) and \(PQ = 15\) cm. (a) Calculate \(QR\). [2] (b) Calculate \(PR\...

The right angle is at \(Q\), so \(PR\) is the hypotenuse. From the \(37^\circ\) angle at \(P\), the side \(QR\) is opposite and \(PQ = 15\) cm is adjacent.

(a) Opposite with adjacent means tangent:

\[ QR = 15 \tan 37^\circ \] [M1]

\[ QR = 15 \times 0.753554\ldots = 11.303\ldots = 11.3 \text{ cm} \] [A1]

(b) Adjacent with hypotenuse means cosine, using the original \(15\) cm rather than the rounded answer to (a):

\[ \cos 37^\circ = \frac{15}{PR} \quad\Rightarrow\quad PR = \frac{15}{\cos 37^\circ} \] [M1]

\[ PR = \frac{15}{0.798635\ldots} = 18.782\ldots = 18.8 \text{ cm} \] [A1]

(c) \(PQ\) and \(QR\) meet at the right angle, so they are the perpendicular base and height:

\[ \text{Area} = \frac{1}{2} \times 15 \times 11.303\ldots \] [M1]

\[ \text{Area} = 84.77\ldots = 84.8 \text{ cm}^2 \] [A1]

Never use the hypotenuse \(PR\) in the area formula: it is not perpendicular to either other side. As a check on part (b), \(\sqrt{15^2 + 11.303^2} = 18.8\) cm, agreeing with the cosine result.