Trigonometry (identities, equations) – A-level Maths exam questions with mark scheme

1CoreCalculator4 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.

2StandardCalculator4 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.

3CoreCalculator3 marks

Solve, for $0\leqslant x<360^\circ$, $$3\sin x=2\cos x,$$ giving your answers to 1 decimal place. [3]

for using $\tan x=\dfrac{\sin x}{\cos x}$ to obtain $\tan x=\tfrac23$M1
$33.7^\circ$ (awrt)A1
$213.7^\circ$ (awrt) – their $33.7^\circ+180^\circ$, and no extra solutions in rangeA1ft
Answer: $x=33.7^\circ,\ 213.7^\circ$
Examiner tip: Divide by $\cos x$, don't square both sides – squaring creates extra wrong solutions you then have to remove.

4StandardCalculator5 marks

Solve, for $0\leqslant\theta<360^\circ$, $$2\sin^{2}\theta+3\cos\theta=3.$$ [5]

for using $\sin^{2}\theta=1-\cos^{2}\theta$M1
for $2\cos^{2}\theta-3\cos\theta+1=0$ (or equivalent 3-term quadratic $=0$)A1
for solving their quadratic in $\cos\theta$: $(2\cos\theta-1)(\cos\theta-1)=0$M1
two of $0^\circ,\ 60^\circ,\ 300^\circ$A1
all three: $0^\circ,\ 60^\circ,\ 300^\circ$ and no others in rangeA1
Answer: $\theta=0^\circ,\ 60^\circ,\ 300^\circ$
Examiner tip: $\cos\theta=1$ gives $\theta=0^\circ$ – it's in the interval because of '$\leqslant$', and it's the solution most often missed.

5HardCalculator4 marks

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

Solve, for $0\leqslant x<180^\circ$, $$\tan(2x-30^\circ)=-\sqrt{3}.$$ [4]

for $2x-30^\circ=120^\circ$ (or $-60^\circ$), i.e. a correct order of operations with the exact valueM1
$x=75^\circ$A1
for adding $180^\circ$ to their $2x-30^\circ$ value to find a second value in the transformed range $-30^\circ\leqslant 2x-30^\circ<330^\circ$M1
$x=165^\circ$ and no others in rangeA1
Answer: $x=75^\circ,\ 165^\circ$
Examiner tip: Change the interval first: $-30^\circ\leqslant 2x-30^\circ<330^\circ$; find all angles there, then add 30 and halve.

6HardCalculator6 marks

(a) Show that the equation $$4\sin x\tan x=15$$ can be written in the form $$4\cos^{2}x+15\cos x-4=0.$$ [3]

(b) Hence solve, for $0\leqslant x<360^\circ$, $$4\sin x\tan x=15,$$ giving your answers to 1 decimal place. [3]

(a) for using $\tan x=\dfrac{\sin x}{\cos x}$ to obtain $\dfrac{4\sin^{2}x}{\cos x}=15$M1
(a) for multiplying by $\cos x$ and using $\sin^{2}x=1-\cos^{2}x$M1
(a) $4-4\cos^{2}x=15\cos x\Rightarrow4\cos^{2}x+15\cos x-4=0$ with no errorsA1*
(b) for solving the quadratic: $(4\cos x-1)(\cos x+4)=0$, giving $\cos x=\tfrac14$M1
(b) $75.5^\circ$ (awrt)A1
(b) $284.5^\circ$ (awrt), with $\cos x=-4$ rejected or ignored and no extra solutionsA1
Answer: (b) $x=75.5^\circ,\ 284.5^\circ$
Examiner tip: State that $\cos x=-4$ has no solutions (since $-1\leqslant\cos x\leqslant1$) rather than leaving it hanging.

7StretchCalculator8 marks

The depth of water, $d$ metres, at a harbour entrance is modelled by $$d=4+1.5\sin(30t)^\circ,\qquad 0\leqslant t\leqslant24,$$ where $t$ is the time in hours after midnight. A sketch of the graph is shown.

(a) State the maximum depth of water predicted by the model. [1]

(b) Find the value of $t$ when this maximum first occurs. [1]

A boat can only pass the entrance when the depth is more than $4.75$ metres.

(d) Find the total length of time between $t=12$ and $t=24$ during which the boat can pass. [2]

(a) $5.5$ mB1
(b) $t=3$B1
(c) for $4+1.5\sin(30t)=4.75\Rightarrow\sin(30t)=0.5$M1
(c) $30t=30^\circ$ and $150^\circ$A1
(c) for dividing their angles by 30dM1
(c) $t=1$ and $t=5$ (1 am and 5 am)A1
(d) for using periodicity (period 12 hours) to find $t=13$ and $t=17$, or $30t=390^\circ,510^\circ$M1
(d) $4$ hoursA1
Answer: (a) $5.5$ m (b) $t=3$ (c) $t=1,\ 5$ (d) $4$ hours
Examiner tip: Multiply out the interval: $0\leqslant t\leqslant12$ means $0\leqslant30t\leqslant360^\circ$, so look for every angle with $\sin=0.5$ up to $360^\circ$.

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