Sine/cosine rules, radians, arcs and sectors, small angles – A-level Maths exam questions with mark scheme

1CoreCalculator5 marks

The diagram shows a sector $OAB$ of a circle with centre $O$ and radius 8 cm. The angle $AOB$ is 1.2 radians. The straight line $AB$ is shown dashed.

(a) Find the length of the arc $AB$. [1]

(b) Find the area of the sector $OAB$. [1]

(c) Find the area of the segment bounded by the arc $AB$ and the chord $AB$, giving your answer to 3 significant figures. [3]

(a) $9.6$ cmB1
(b) $38.4\ \text{cm}^{2}$B1
(c) Area of triangle $=\tfrac12\times8^{2}\times\sin1.2\ (=29.82\ldots)$ – calculator in radian modeM1
Sector area $-$ triangle areaM1
$8.57\ \text{cm}^{2}$A1
Answer: (a) $9.6$ cm (b) $38.4\ \text{cm}^2$ (c) $8.57\ \text{cm}^2$
Examiner tip: Make sure your calculator is in radians for $\sin1.2$ – degree mode gives a totally wrong triangle area.

2CoreCalculator3 marks

Given that $\theta$ is small and measured in radians, use the small angle approximations for $\sin\theta$ and $\cos\theta$ to find an approximate value of $$\frac{1-\cos4\theta}{2\theta\sin3\theta}$$ [3]

Uses $\cos4\theta\approx1-\dfrac{(4\theta)^{2}}{2}=1-8\theta^{2}$M1
Uses $\sin3\theta\approx3\theta$ to give $\dfrac{8\theta^{2}}{6\theta^{2}}$M1
$\dfrac43$A1
Answer: $\dfrac43$
Examiner tip: Square the whole angle: $\dfrac{(4\theta)^{2}}{2}=8\theta^{2}$, not $\dfrac{4\theta^{2}}{2}$.

3CoreCalculator4 marks

The diagram shows triangle $ABC$ with $AB=7\text{ cm}$, $AC=9\text{ cm}$ and angle $BAC=50^\circ$.

(a) Find the length of $BC$, giving your answer in cm to 3 significant figures. [2]

(b) Find the area of triangle $ABC$, giving your answer in $\text{cm}^{2}$ to 3 significant figures. [2]

(a) for $BC^{2}=7^{2}+9^{2}-2\times7\times9\cos50^\circ$M1
(a) $BC=7.00\text{ cm}$ (awrt 7.00)A1
(b) for $\tfrac12\times7\times9\times\sin50^\circ$M1
(b) $24.1\text{ cm}^{2}$ (awrt)A1
Answer: (a) $7.00\text{ cm}$ (b) $24.1\text{ cm}^{2}$
Examiner tip: Check your calculator is in degrees – in radians $\cos 50$ gives a nonsense length.

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4StandardCalculator4 marks

In triangle $ABC$, $AB=10\text{ cm}$, $BC=7\text{ cm}$ and angle $BAC=40^\circ$.

(a) Find the two possible values of angle $ACB$, giving your answers in degrees to 1 decimal place. [3]

Given that angle $ACB$ is obtuse,

(b) find the size of angle $ABC$. [1]

(a) for the sine rule: $\dfrac{\sin C}{10}=\dfrac{\sin40^\circ}{7}$M1
(a) $66.7^\circ$ (awrt)A1
(a) $113.3^\circ$ (awrt), i.e. $180^\circ-$ their acute angleA1ft
(b) $26.7^\circ$ (awrt) – follow through $140^\circ-$ their obtuse angleB1ft
Answer: (a) $66.7^\circ$ or $113.3^\circ$ (b) $26.7^\circ$
Examiner tip: When the sine rule gives an angle, always check whether $180^\circ-\theta$ also works – that's the ambiguous case.

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