A shop sign is a right-angled triangle \(LMN\) with angle \(LMN = 90^\circ\), \(LM = 24\) cm and angle \(MLN = 53^\circ\). (a) Calculate \(MN\). [2] (b) Cal...

The sign is a right-angled triangle with the right angle at \(M\), so \(LM\) and \(MN\) are the two shorter sides (the legs) and \(LN\) is the hypotenuse. Working from the \(53^\circ\) angle at \(L\): \(LM\) is the side adjacent to it, \(MN\) is the side opposite it, and \(LN\) is the hypotenuse. Choosing the right trigonometric ratio is simply a matter of naming those three sides correctly first.

(a) \(MN\) is opposite and \(LM = 24\) cm is adjacent, so use the tangent ratio:

\[\tan 53^\circ = \frac{MN}{24} \quad\Rightarrow\quad MN = 24\tan 53^\circ\]

That substitution earns [M1]. Evaluating, \(24 \times 1.32704\ldots = 31.849\ldots\), so \(MN = 31.8\) cm to 3 significant figures [A1].

(b) \(LN\) is the hypotenuse and \(LM = 24\) cm is adjacent, so use cosine:

\[\cos 53^\circ = \frac{24}{LN} \quad\Rightarrow\quad LN = \frac{24}{\cos 53^\circ}\]

Rearranging to put \(LN\) on top scores [M1]. Then \(LN = 24 \div 0.601815\ldots = 39.879\ldots\), so \(LN = 39.9\) cm [A1]. A useful check: the hypotenuse must be the longest side, and \(39.9 \gt 31.8 \gt 24\), which it is.

(c) Because the angle at \(M\) is \(90^\circ\), the two legs are already perpendicular, so they act as base and height directly and no extra trigonometry is needed:

\[\text{Area} = \tfrac{1}{2} \times 24 \times 31.849\ldots = 382.1\ldots\]

giving \(382\) cm\(^2\) [B1], and this mark is awarded ft (follow through) on the candidate's own value of \(MN\).

Note the unrounded \(31.849\) is used in part (c), not the rounded \(31.8\). Rounding at every stage and then rounding again is the commonest source of a lost accuracy mark: keep full calculator accuracy in the working and round only the final answer to 3 significant figures.