Right-angled trigonometry & exact values – GCSE Maths exam questions with mark scheme

1Grade 4–5Calculator2 marks

The diagram shows a right-angled triangle. The hypotenuse is 12 cm and one angle is 35°.

Work out the length $x$. Give your answer correct to 3 significant figures. [2]

$\sin35^\circ=\dfrac{x}{12}$ or $x=12\sin35^\circ$M1
6.88 cm (accept 6.882 to 6.883)A1
Answer: 6.88 cm
Examiner tip: Label the sides O, A, H from the given angle first – $x$ is opposite 35°, so use sine.

2Grade 4–5Calculator2 marks

In the right-angled triangle shown, the side opposite angle $\theta$ is 7 cm and the side adjacent to $\theta$ is 10 cm.

Work out the size of angle $\theta$. Give your answer correct to 1 decimal place. [2]

$\tan\theta=\dfrac{7}{10}$M1
35.0° (accept 34.99…)A1
Answer: 35.0°
Examiner tip: Use the inverse function ($\tan^{-1}$) to find an angle, and make sure your calculator is in degrees mode.

3Grade 4–5Calculator3 marks

The diagram shows a right-angled triangle. The angle at the top vertex is 52° and the side between this angle and the right angle is 8.5 cm.

Work out the length of the hypotenuse, $y$. Give your answer correct to 3 significant figures. [3]

$\cos52^\circ=\dfrac{8.5}{y}$M1
$y=\dfrac{8.5}{\cos52^\circ}$M1
13.8 cm (accept 13.80 to 13.81)A1
Answer: 13.8 cm
Examiner tip: When the unknown is on the bottom of the fraction, divide – $8.5\times\cos52^\circ$ is a very common wrong step.

4Grade 6–7HigherNon-calculator5 marks

(a) Write down the exact value of
(i) $\sin30^\circ$ (ii) $\tan45^\circ$ (iii) $\cos30^\circ$ [2]

The diagram shows a right-angled triangle with hypotenuse 14 cm and an angle of 30°.

(b) Work out the exact length of $a$. [1]

(a) at least two of $\tfrac12,\;1,\;\tfrac{\sqrt3}{2}$ correctB1
all three correct: $\tfrac12,\;1,\;\tfrac{\sqrt3}{2}$B1
(b) $a=14\times\tfrac12=7$ cmB1
(c) $b=14\cos30^\circ=14\times\tfrac{\sqrt3}{2}$ (or $\sqrt{14^2-7^2}=\sqrt{147}$)M1
$7\sqrt3$ cm caoA1
Answer: (a) $\tfrac12,\;1,\;\tfrac{\sqrt3}{2}$ (b) 7 cm (c) $7\sqrt3$ cm
Examiner tip: 'Exact' means leave the surd – a decimal like 12.12 scores M1 at most.

5Grade 4–5Calculator3 marks

Sam stands on level ground 40 m from the base of a vertical tower. His eyes are 1.6 m above the ground. The angle of elevation of the top of the tower from his eyes is 28°.

Work out the height of the tower. Give your answer correct to 1 decimal place. [3]

$\tan28^\circ=\dfrac{h}{40}$ or $40\tan28^\circ$M1
$40\tan28^\circ+1.6$ (21.26… + 1.6)M1
22.9 m (accept 22.86 to 22.87)A1
Answer: 22.9 m
Examiner tip: Add Sam's eye height at the end – the triangle only goes up from eye level.

6Grade 6–7HigherCalculator4 marks

ADC is a straight line. BD is perpendicular to AC. AB = 14 cm, angle BAD = 40° and angle BCD = 62°.

Work out the length of BC. Give your answer correct to 3 significant figures. [4]

$BD=14\sin40^\circ$M1
BD = 8.999… (accept 9.00)A1
$\sin62^\circ=\dfrac{BD}{BC}$ or $BC=\dfrac{\text{their }BD}{\sin62^\circ}$M1
10.2 cm (accept 10.19)A1
Answer: 10.2 cm
Examiner tip: Work out the shared side BD first and keep it unrounded in your calculator before using the second triangle.

7Grade 8–9HigherCalculator4 marks

ABCDEFGH is a cuboid. AB = 8 cm, BC = 6 cm and CG = 5 cm.

Work out the size of the angle between the diagonal AG and the base ABCD. Give your answer correct to 1 decimal place. [4]

$AC^2=8^2+6^2$ or $AC=\sqrt{8^2+6^2}$ (finds the diagonal of the base) or $AG^2=8^2+6^2+5^2$ (3D Pythagoras)M1
$AC=10$ or $AG=\sqrt{125}=11.18\ldots$A1
$\tan GAC=\dfrac{5}{\text{their }AC}$ or $\sin GAC=\dfrac{5}{\text{their }AG}$ or $\cos GAC=\dfrac{\text{their }AC}{\text{their }AG}$M1
26.6° (accept 26.5 to 26.6)A1
Answer: 26.6°
Examiner tip: The angle with the base is between AG and its projection AC – find AC first, then use the right angle at C.

8Grade 1–3FoundationNon-calculator3 marks

The diagram shows a right-angled triangle ABC. AB = 12 cm, BC = 9 cm and AC = 15 cm. Angle BAC = $\theta$.

(a) Write down the name of the side that is the hypotenuse. [1]

(b) Write $\tan\theta$ as a fraction in its simplest form. [2]

(a) AC (accept CA, or '15 cm' clearly identified)B1
(b) $\dfrac{9}{12}$ (opposite ÷ adjacent seen) — allow $\tan\theta=\dfrac{9}{12}$M1
$\dfrac34$ cao (do not accept 0.75 for this mark)A1
Answer: (a) AC (b) $\tfrac34$
Examiner tip: Label opp, adj and hyp from the angle θ first – the 15 cm side is a distractor for tan.

9Grade 4–5Calculator3 marks

A ladder of length 6.5 m leans against a vertical wall on horizontal ground. The foot of the ladder is 1.8 m from the bottom of the wall.

Safety guidance says the angle between a ladder and the ground should be between 70° and 80°.

Is this ladder positioned safely? You must show your working. [3]

$\cos x=\dfrac{1.8}{6.5}$ (or full method via Pythagoras then $\sin$/$\tan$, e.g. $\tan x=\dfrac{\sqrt{6.5^2-1.8^2}}{1.8}$)M1
$x=73.9^\circ$ (accept 73.9–74)A1
Yes, it is safe, since their angle lies between 70° and 80° (ft their angle if M1 awarded)C1
Answer: 73.9°, which is between 70° and 80°, so the ladder is safe.
Examiner tip: A bare 'yes' scores nothing – the C1 needs the angle and a comparison with 70° and 80°.

10Grade 4–5Calculator3 marks

T is the top of a vertical cliff, 85 m above sea level. From T, the angle of depression of a boat is 17°.

Work out the horizontal distance, $d$, from the bottom of the cliff to the boat. Give your answer correct to the nearest metre. [3]

$\tan17^\circ=\dfrac{85}{d}$ (or uses angle 73° at T: $\tan73^\circ=\dfrac{d}{85}$)M1
$d=\dfrac{85}{\tan17^\circ}$ (or $85\tan73^\circ$)M1
278 m (accept 278.0…)A1
Answer: 278 m
Examiner tip: The angle of depression is measured down from the horizontal, so it equals the angle of elevation at the boat (alternate angles) – don't put 17° at the top of the cliff wall.

11Grade 4–5Non-calculator4 marks

(a) The table shows exact values of $\sin\theta$, $\cos\theta$ and $\tan\theta$. Five values are missing.

$\theta$	0°	30°	45°	60°	90°
$\sin\theta$	0	$\tfrac12$	$\tfrac{\sqrt2}{2}$	$\tfrac{\sqrt3}{2}$	(i)
$\cos\theta$	1	$\tfrac{\sqrt3}{2}$	(ii)	(iii)	0
$\tan\theta$	(iv)	(v)	1	$\sqrt3$	—

Write down the exact values for (i), (ii), (iii), (iv) and (v). [2]

(b) Mia says, "$\sin60^\circ$ is double $\sin30^\circ$, because 60 is double 30."

Is Mia correct? You must show how you get your answer. [2]

(a) at least 3 of the 5 values correct: (i) 1, (ii) $\tfrac{\sqrt2}{2}$, (iii) $\tfrac12$, (iv) 0, (v) $\tfrac{\sqrt3}{3}$B1
all 5 correct — accept $\tfrac{1}{\sqrt2}$ for (ii) and $\tfrac{1}{\sqrt3}$ for (v); decimals are not exact and score B0 for that valueB1
(b) $\sin60^\circ=\dfrac{\sqrt3}{2}$ and $2\times\sin30^\circ=2\times\dfrac12=1$ (both values seen)M1
No, with correct comparison, e.g. $\dfrac{\sqrt3}{2}\ne1$ (accept 0.866… ≠ 1)C1
Answer: (a) (i) 1 (ii) $\tfrac{\sqrt2}{2}$ (iii) $\tfrac12$ (iv) 0 (v) $\tfrac{\sqrt3}{3}$ (b) No: $\sin60^\circ=\tfrac{\sqrt3}{2}$ but $2\sin30^\circ=1$
Examiner tip: In (b) the mark is for the two values side by side – 'No, it doesn't work like that' with no numbers scores 0.

12Grade 6–7HigherNon-calculator5 marks

ABC and ACD are right-angled triangles. Angle ABC = 90° and angle ACD = 90°.
AB = 6 cm, angle ACB = 60° and angle CAD = 45°.

Show that $AD=4\sqrt6$ cm. [5]

$\sin60^\circ=\dfrac{6}{AC}$ or $AC=\dfrac{6}{\sin60^\circ}$ (or $\tan60^\circ=\dfrac{6}{BC}$ as a first step towards $AC$ via Pythagoras)M1
$AC=6\div\dfrac{\sqrt3}{2}\;\left(=\dfrac{12}{\sqrt3}\right)$ — exact value of $\sin60^\circ$ used (or $BC=\dfrac{6}{\sqrt3}=2\sqrt3$ and $AC=\sqrt{6^2+(2\sqrt3)^2}$)M1
$AC=4\sqrt3$ (oe exact, e.g. $\dfrac{12}{\sqrt3}$ or $\sqrt{48}$)A1
$\cos45^\circ=\dfrac{AC}{AD}$, i.e. $AD=4\sqrt3\div\dfrac{\sqrt2}{2}$ (or $CD=AC$ as triangle ACD is isosceles, then $AD=\sqrt{48+48}$)M1
$AD=\dfrac{8\sqrt3}{\sqrt2}=4\sqrt6$ shown with no errors (or $\sqrt{96}=\sqrt{16\times6}=4\sqrt6$)A1
Answer: $AC=4\sqrt3$ cm, so $AD=4\sqrt3\div\tfrac{\sqrt2}{2}=4\sqrt6$ cm
Examiner tip: In a 'show that' every step must be exact – a decimal such as 9.798 never proves $4\sqrt6$ and loses the final A1.

13Grade 6–7HigherCalculator4 marks

A plane flies 210 km due east from airport A to point B. It then flies 130 km due south from B to point C.

Work out the bearing of A from C. Give your answer correct to the nearest degree. [4]

$\tan x=\dfrac{210}{130}$ (angle at C) or $\tan y=\dfrac{130}{210}$ (angle at A)M1
$x=58.2^\circ$ or $y=31.8^\circ$A1
$360-58.2$ or $270+31.8$ (correct method for the bearing from their angle)M1
302° (accept 301.8°)A1
Answer: 302°
Examiner tip: Draw a north line at C – the bearing of A from C is measured clockwise from north at C, so 58° on its own is the angle, not the bearing.

14Grade 8–9HigherCalculator4 marks

VABCD is a pyramid with a square base ABCD of side 10 cm. The vertex V is vertically above the centre of the base.
VA = VB = VC = VD = 13 cm.

Work out the size of the angle between the edge VA and the base ABCD. Give your answer correct to 1 decimal place. [4]

$AC=\sqrt{10^2+10^2}\;(=\sqrt{200}=14.14\ldots)$M1
$AM=\tfrac12AC\;(=7.07\ldots)$ where M is the centre of the baseM1
$\cos\theta=\dfrac{7.07\ldots}{13}$ (or height $VM=\sqrt{13^2-7.07^2}=10.9\ldots$ then $\sin\theta=\dfrac{10.9}{13}$ or $\tan\theta=\dfrac{10.9}{7.07}$)M1
57.0° (accept 57.0–57.1)A1
Answer: 57.0°
Examiner tip: The angle with the base sits in triangle VAM, using HALF the diagonal – using 10 cm or the full diagonal 14.1 cm is the most common error.

15Grade 8–9HigherNon-calculator5 marks

The diagram shows a right-angled triangle. The side adjacent to the 30° angle is $x$ cm and the hypotenuse is $(x+2)$ cm.

Show that $x=6+4\sqrt3$. [5]

$\cos30^\circ=\dfrac{x}{x+2}$M1
$\dfrac{x}{x+2}=\dfrac{\sqrt3}{2}$ leading to $2x=\sqrt3x+2\sqrt3$ (or $2x=\sqrt3(x+2)$)A1
collects $x$ terms and factorises: $x(2-\sqrt3)=2\sqrt3$, so $x=\dfrac{2\sqrt3}{2-\sqrt3}$M1
multiplies numerator and denominator by $(2+\sqrt3)$M1
$x=\dfrac{4\sqrt3+6}{4-3}=6+4\sqrt3$ with denominator $=1$ clearly shownA1
Answer: $x=\dfrac{2\sqrt3}{2-\sqrt3}=\dfrac{2\sqrt3(2+\sqrt3)}{1}=6+4\sqrt3$
Examiner tip: Students usually stall at $x(2-\sqrt3)=2\sqrt3$ – you must rationalise using the conjugate $(2+\sqrt3)$, not just $\sqrt3$.

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