Secondary 3 E Math · Trigonometry

Cosine rule, sine rule and area of a non-right-angled triangle

The question

In triangle $ABC$ , $AB = 9.4\ \text{cm}$ , $BC = 12.7\ \text{cm}$ and $\angle ABC = 108^\circ$ .

(a) Calculate $AC$ . (3 marks)

(b) Calculate $\angle BAC$ . (3 marks)

Give each answer correct to 3 significant figures.

The solution, line by line

01M1
(a) Two sides and the angle between them are given, so use the cosine rule with $AC$ opposite the known angle $B$ :
$AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos B$
Why this earns the mark
Choosing the rule is the first decision an examiner looks for. The sine rule needs a side paired with its opposite angle, and there is no such pair here — SAS with the angle sandwiched between the two sides is always the cosine rule.
02M1
$AC^2 = 9.4^2 + 12.7^2 - 2(9.4)(12.7)\cos 108^\circ$ $AC^2 = 88.36 + 161.29 + 73.779\ldots = 323.42\ldots$
Why this earns the mark
Substitute, then let the calculator finish. Because $108^\circ$ is obtuse, $\cos 108^\circ$ is negative, so the last term adds. Sanity check: an obtuse angle at $B$ must make $AC$ longer than either given side, and it does.
03A1
$AC = \sqrt{323.42\ldots} = 17.983\ldots \approx 18.0\ \text{cm}$
Why this earns the mark
Accuracy mark. Take the square root — forgetting it leaves $AC^2$ on the answer line. Round only here, to 3 s.f. as asked, and keep the unrounded value in the calculator for part (b).
04M1
(b) Now a side and its opposite angle are known ( $AC$ opposite $B$ ), so the sine rule applies:
$\frac{\sin \angle BAC}{BC} = \frac{\sin B}{AC}$
Why this earns the mark
Pair each side with the angle facing it. $\angle BAC$ faces $BC$ , and $B$ faces $AC$ — mismatching the pairs is the classic sine-rule error and loses the method mark even when the arithmetic is faultless.
05M1
$\sin \angle BAC = \frac{12.7 \sin 108^\circ}{17.983\ldots} = 0.67161\ldots$
Why this earns the mark
Use the unrounded $AC$ , not $18.0$ . Rounding twice is what turns a correct method into an answer that misses the mark scheme’s tolerance.
06A1
$\angle BAC = \sin^{-1}(0.67161\ldots) = 42.19\ldots^\circ \approx 42.2^\circ$
Why this earns the mark
Accuracy mark. The ambiguous case does not bite here: a triangle can hold only one obtuse angle and $B = 108^\circ$ has already used it, so the acute value from the calculator is the only possibility. Check: $42.2^\circ + 108^\circ < 180^\circ$ . ✓
07M1
(c) Use the included-angle area formula:
$\text{Area} = \tfrac{1}{2}(AB)(BC)\sin B = \tfrac{1}{2}(9.4)(12.7)\sin 108^\circ$
Why this earns the mark
Use the two given sides and the angle between them, not the values you calculated — this way part (c) cannot inherit an error from parts (a) or (b). $\tfrac12 \times \text{base} \times \text{height}$ does not apply: $9.4$ and $12.7$ are not perpendicular.
08A1
$\text{Area} = 56.76\ldots \approx 56.8\ \text{cm}^2$
Why this earns the mark
Accuracy mark, and the unit changes to $\text{cm}^2$ — an area written in $\text{cm}$ is routinely penalised.

Final answer

(a) $AC = 18.0\ \text{cm}$

(b) $\angle BAC = 42.2^\circ$

(c) Area $= 56.8\ \text{cm}^2$

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