Pythagoras & right-angled trigonometry in 2D – IGCSE Maths exam questions with mark scheme

1Grade 4–5Calculator3 marks

$ABC$ is a right-angled triangle.

$AB = 6.8$ cm and $BC = 11.5$ cm. Angle $ABC = 90^\circ$.

Calculate the length of $AC$, marked $x$ on the diagram.

Give your answer correct to 3 significant figures. [3]

for $6.8^2 + 11.5^2$ or $46.24 + 132.25$ or $178.49$ seenM1
for $\sqrt{6.8^2 + 11.5^2}$ or $\sqrt{178.49}$M1
$13.4$ (cm) accept answers rounding to $13.4$, e.g. $13.36$; if no working, $13.4$ scores all 3 marksA1
Answer: $13.4$ cm
Examiner tip: Square, ADD, then square root for the hypotenuse; subtracting gives $9.27$, which is shorter than $11.5$ cm and cannot be the longest side.

2Grade 4–5Calculator3 marks

$PQR$ is a right-angled triangle.

$PR = 15$ cm. Angle $RPQ = 34^\circ$. Angle $PQR = 90^\circ$.

Calculate the length of $PQ$, marked $x$ on the diagram.

Give your answer correct to 1 decimal place. [3]

for recognising cosine is needed, e.g. $\cos 34^\circ = \dfrac{x}{15}$M1
for $x = 15 \times \cos 34^\circ$M1
$12.4$ (cm) accept $12.43$ to $12.44$; an answer from a calculator in radian mode (e.g. $-12.7$) scores M2 A0 if the method is shownA1
Answer: $12.4$ cm
Examiner tip: $x$ is next to the $34^\circ$ angle, so it is the adjacent side: use cos, not sin (sin gives $8.4$ cm, which is the side $QR$).

3Grade 4–5Calculator3 marks

Point $A$ has coordinates $(-3,\ 2)$.

Point $B$ has coordinates $(5,\ -4)$.

Work out the length of the line segment $AB$. [3]

for the horizontal and vertical distances, $8$ and $6$ (or $-8$ and $6$, etc.) seen or usedM1
for $8^2 + 6^2$ or $64 + 36$ or $\sqrt{100}$M1
$10$ (units not required)A1
Answer: $10$
Examiner tip: The horizontal distance from $-3$ to $5$ is $8$, not $2$: count across zero, or work out $5 - (-3)$.

Stuck on Pythagoras & right-angled trigonometry in 2D?

Get it sorted in a free 30-minute session

We work through your weakest questions on this topic on Zoom. I reply the same day.

Free 30-min session on this topic Send me 5 questions a day

4Grade 4–5Calculator3 marks

(a) A triangle has sides of length 20 cm, 48 cm and 52 cm.

Show that the triangle is a right-angled triangle. [2]

(b) Sam says,

“A triangle with sides of length 5 cm, 9 cm and 10 cm is right-angled, because $5 + 9 > 10$.”

Is Sam correct? Give a reason for your answer. [1]

(a) for $20^2 + 48^2$ ( $= 400 + 2304$ ) and $52^2$ evaluated or comparedM1
$400 + 2304 = 2704$ and $52^2 = 2704$, so by Pythagoras (converse) the triangle is right-angled both values $2704$ and a conclusion neededA1
(b) No, with reason, e.g. $5^2 + 9^2 = 106$ but $10^2 = 100$ and $106 \ne 100$ ignore any comment about $5 + 9 > 10$ (that only shows a triangle can be drawn)C1
Answer: (a) $20^2 + 48^2 = 2704 = 52^2$ (b) No: $5^2 + 9^2 = 106 \ne 100 = 10^2$
Examiner tip: In a 'show that', write both totals and a conclusion; just writing $52^2 = 2704$ without the comparison is not enough for the A mark.

5Grade 4–5Calculator4 marks

The diagram shows a wheelchair ramp leading up to a doorway.

The sloping surface of the ramp is 4.6 m long.

The top of the ramp is 0.42 m above the ground.

Building rules say that the angle between a ramp and the horizontal ground must be no more than $5^\circ$.

Does this ramp meet the building rules?

You must show all your working. [4]

for $\sin\theta = \dfrac{0.42}{4.6}$M1
for $\theta = \sin^{-1}\left(\dfrac{0.42}{4.6}\right)$M1
$5.2(38...)^\circ$ accept $5.24^\circ$ or $5.2^\circ$A1
No, it does not meet the rules, since $5.24^\circ > 5^\circ$ dependent on M2; follow through their angle from a correct trig method; allow comparison of $0.42/4.6 = 0.0913$ with $\sin 5^\circ = 0.0872$ for full marksC1
Answer: No: the angle is $5.24^\circ$, which is more than $5^\circ$
Examiner tip: 4.6 m is the sloping side (hypotenuse), so this needs sin; using tan gives $5.22^\circ$, which still exceeds $5^\circ$ but loses the method marks.

6Grade 4–5Calculator4 marks

An aircraft flies from airport $P$ on a bearing of $125^\circ$.

It flies 280 km in a straight line and reaches point $Q$.

Work out how far $Q$ is

(a) east of $P$, [2]

(b) south of $P$. [2]

Give each answer correct to 3 significant figures.

(a) for $280 \times \sin 55^\circ$ or $280 \times \cos 35^\circ$ (angle with south line is $180^\circ - 125^\circ = 55^\circ$)M1
$229$ (km) accept $229.3$ to $229.4$A1
(b) for $280 \times \cos 55^\circ$ or $280 \times \sin 35^\circ$, or $\sqrt{280^2 - 229.36^2}$M1
$161$ (km) accept $160.6$ to $160.7$A1
Answer: (a) $229$ km (b) $161$ km
Examiner tip: Draw a north–south line at $P$ and complete the right-angled triangle first; the angle inside it is $55^\circ$ (to south) or $35^\circ$ (to east), never $125^\circ$.

7Grade 4–5Calculator5 marks

$ABC$ is an isosceles triangle.

$AB = AC = 9.4$ cm and $BC = 7.6$ cm.

(a) Work out the area of triangle $ABC$.

Give your answer correct to 3 significant figures. [3]

(b) Work out the size of angle $ABC$.

Give your answer correct to 1 decimal place. [2]

(a) for halving the base and using Pythagoras, e.g. $9.4^2 - 3.8^2$ ( $= 73.92$ )M1
for $\dfrac{1}{2} \times 7.6 \times \sqrt{73.92}$ or $\dfrac12 \times 7.6 \times 8.59(7...)$M1
$32.7$ (cm$^2$) accept $32.6$ to $32.7$A1
(b) for $\cos B = \dfrac{3.8}{9.4}$ or $\tan B = \dfrac{8.59...}{3.8}$ or $\sin B = \dfrac{8.59...}{9.4}$M1
$66.2^\circ$ accept $66.1^\circ$ to $66.2^\circ$A1
Answer: (a) $32.7$ cm$^2$ (b) $66.2^\circ$
Examiner tip: Split the isosceles triangle down the middle: the right-angled triangle has base $3.8$ cm, not $7.6$ cm.

8Grade 6–7HigherCalculator4 marks

$ABC$ is a right-angled triangle. Angle $ACB = 90^\circ$.

$D$ is a point on $BC$.

$AC = 6$ cm. Angle $CAD = 25^\circ$. Angle $DAB = 33^\circ$.

Work out the length of $BD$.

Give your answer correct to 3 significant figures. [4]

for $BC = 6\tan 58^\circ$ ( $= 9.60(2...)$ ) (angle $CAB = 25^\circ + 33^\circ = 58^\circ$)M1
for $DC = 6\tan 25^\circ$ ( $= 2.79(78...)$ )M1
for $6\tan 58^\circ - 6\tan 25^\circ$M1
$6.80$ (cm) accept $6.8$ or answers in the range $6.80$ to $6.81$A1
Answer: $6.80$ cm
Examiner tip: Angle $DAB$ is not in a right-angled triangle, so $6\tan 33^\circ = 3.90$ is wrong: find $BC$ and $DC$ separately, then subtract.

9Grade 6–7HigherCalculator4 marks

$TF$ is a vertical cliff of height 85 m. $F$ is at sea level.

$A$ and $B$ are two boats on the sea, in a straight line with $F$.

The angle of depression of $A$ from $T$ is $18^\circ$.

The angle of depression of $B$ from $T$ is $31^\circ$.

Work out the distance $AB$.

Give your answer correct to the nearest metre. [4]

for $FA = \dfrac{85}{\tan 18^\circ}$ ( $= 261.6...$ ) or $85\tan 72^\circ$M1
for $FB = \dfrac{85}{\tan 31^\circ}$ ( $= 141.4(6...)$ ) or $85\tan 59^\circ$M1
for $FA - FB$, e.g. $261.6 - 141.5$M1
$120$ (m) accept $120.1$; must be correct to nearest metre from correct workingA1
Answer: $120$ m
Examiner tip: The angle of depression is measured down from the horizontal at $T$, so it equals the angle of elevation at the boat: use $\tan 18^\circ = 85/FA$, not $85 \times \tan 18^\circ$.

10Grade 8–9HigherCalculator6 marks

The diagram shows a right-angled triangle.

The lengths of the sides, in centimetres, are $x$, $x + 7$ and $x + 8$.

(a) Show that $x^2 - 2x - 15 = 0$ [2]

(b) Hence find the size of the smallest angle in the triangle.

Give your answer correct to 1 decimal place. [4]

(a) for $x^2 + (x+7)^2 = (x+8)^2$M1
for correct expansion $x^2 + x^2 + 14x + 49 = x^2 + 16x + 64$ leading to $x^2 - 2x - 15 = 0$ with no errors seen (answer given: every step must be shown)A1
(b) for $(x - 5)(x + 3) = 0$ or correct use of the quadratic formulaM1
$x = 5$ (rejecting $x = -3$ as a length cannot be negative) sides $5$, $12$, $13$A1
for a correct trig ratio for the angle opposite the side of length $x$, e.g. $\tan\theta = \dfrac{5}{12}$ or $\sin\theta = \dfrac{5}{13}$M1
$22.6^\circ$A1
Answer: (a) shown (b) $22.6^\circ$
Examiner tip: The smallest angle is opposite the shortest side ($5$ cm), so use $\tan^{-1}(5/12)$; $\tan^{-1}(12/5) = 67.4^\circ$ is the other acute angle.

Stuck on this topic?

Book a free 30-minute session on WhatsApp and we'll find exactly which step is costing you the marks. Or build a timed paper on it.

Free 30-minute session