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Sec 4 E Maths: Trigonometry and Bearings, practice questions & worked solutions

Trigonometry and bearings questions selected from 2023 to 2025 Singapore school E Maths prelim papers, each with the school’s full worked solution.

10 questionsStep-by-step solutionsFree to access

About this topic & key methods

Trigonometry in Sec 3 and Sec 4 E Maths moves from right-angled triangles to the sine and cosine rules, bearings, angles of elevation and depression, and simple 3D problems. Paper 1 items here are short; Paper 2 items combine several steps, often with bearings and a 3D diagram.

Attempt each question on paper before opening its worked solution. The solutions are the schools’ own marking schemes, so some steps are brief.

Key methods for trigonometry and bearings

In a right-angled triangle, use sine, cosine and tangent ratios, and Pythagoras’ theorem for the third side.
Sine rule $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}$ when you know a side and its opposite angle. Since $\sin(180^\circ-\theta)=\sin\theta$ , check whether the angle can be obtuse.
Cosine rule $a^2=b^2+c^2-2bc\cos A$ for two sides and the included angle, or for all three sides.
Area of a triangle $=\tfrac12 ab\sin C$ .
Bearings are measured clockwise from North and written with three figures. Parallel North lines give co-interior angles that add up to $180^\circ$ .
The greatest angle of elevation or depression from a path occurs at the shortest (perpendicular) distance to it.
In 3D problems, find the right-angled triangle that contains the angle, and work out its base on the horizontal plane first.

For lessons that build these methods week by week, see our E Maths tuition for Sec 3 and Sec 4.

Trigonometry and Bearings questions with worked solutions

Unless the question specifies otherwise, give non-exact answers to 3 significant figures and angles in degrees to 1 decimal place.

Q1Yuying Secondary School2025Prelim P1 Q3●○○ Foundational3 marks

Sine of an obtuse angle and radians, Yuying Secondary School, 2025 Prelim P1 Q3

Given that $0<\theta<180^\circ$ , find the values of $\theta$ for $\sin\theta=0.6453$ . [2]
Express $240^\circ$ in radians, in terms of $\pi$ . [1]

Show worked solution▾

(a)

$40.2^\circ$ (1 dp) or $139.8^\circ$ (1 dp)

(b)

$\frac\pi{180}\times240=\frac{4\pi}{3}\text{ rad}$

Answer: (a) $40.2^\circ$ , $139.8^\circ$ (b) $\dfrac{4\pi}{3}$ rad

Q2Naval Base Secondary School2024Prelim P1 Q20●●○ Standard4 marks

Trigonometric ratios in a right-angled triangle, Naval Base Secondary School, 2024 Prelim P1 Q20

Naval Base Secondary School 2024 P1 Q20: diagram for the question

In triangle $ABC$ , angle $ABC=90^\circ$ , $BX=2XC$ and $\sin\angle AXC=\dfrac35$ .

Express as a fraction in its lowest terms,

$\tan\angle AXB$ , [2]
$\tan\angle ACB$ . [2]

Show worked solution▾

(a)

$\begin{aligned}\sin\angle AXC&=\dfrac35\\BX&=4\\\tan\angle AXB&=\dfrac34\end{aligned}$

(b)

$\begin{aligned}\tan\angle ACB&=\dfrac3{4+2}\\&=\dfrac12\end{aligned}$

Answer: (a) $\dfrac34$ (b) $\dfrac12$

Q3St Joseph’s Institution2024Prelim P1 Q8●●○ Standard3 marks

Sine rule: the ambiguous case, St Joseph’s Institution, 2024 Prelim P1 Q8

In triangle $XYZ$ , $YZ=13$ cm, $XY=20$ cm and angle $YXZ=35^\circ$ .

Find the two possible values of angle $XZY$ . [3]

Show worked solution▾

By Sine Rule

$\begin{aligned}\dfrac{\sin(\angle YXZ)}{13}&=\dfrac{\sin(\angle XZY)}{20}\\[4pt]\sin(\angle XZY)&=\dfrac{20\sin(35^\circ)}{13}\\[4pt]\angle XZY&=\sin^{-1}(0.882425)\\\angle XZY&=61.936^\circ\\\angle XZY&=61.9^\circ\end{aligned}$

$\begin{aligned}\angle XZY&=180^\circ-61.936^\circ\text{ (By obtuse angle ratio)}\\\angle XZY&=118.1^\circ\text{ (By obtuse angle ratio)}\end{aligned}$

Answer: $61.9^\circ$ or $118.1^\circ$

Q4CHIJ St. Nicholas Girls’ School2023Prelim P1 Q17●●○ Standard5 marks

Sine rule and area of a triangle, CHIJ St. Nicholas Girls’ School, 2023 Prelim P1 Q17

CHIJ St. Nicholas Girls’ School 2023 P1 Q17: diagram for the question

In the triangle $ABC$ , $AB=12.5$ cm, angle $ABC=40\degree$ , angle $ACB=55\degree$ and $AC=AD$ .

Calculate $AC$ . [2]
Calculate the area of triangle $ADB$ . [3]

Show worked solution▾

(a)

$\begin{aligned}\dfrac{AC}{\sin40\degree}&=\dfrac{12.5}{\sin55\degree}\\AC&=\dfrac{12.5\sin40\degree}{\sin55\degree}\\AC&=9.808\\AC&=9.81\text{ cm (3sf)}\end{aligned}$

(b)

$\triangle ACD$ is isosceles with $\angle ACD=\angle ADC$

$\angle ADC=55\degree$

$\therefore\angle ADB=180\degree-55\degree=125\degree$ (adj $\angle s$ on a str line)

$\therefore\angle DAB=180\degree-125\degree-40\degree=15\degree$ ( $\angle$ sum of $\Delta$ )

or $\therefore\angle DAB=55\degree-40\degree=15\degree$ (ext $\angle$ of $\Delta$ )

$\begin{aligned}\text{Area of }\triangle ADB&=\dfrac12(9.808)(12.5)\sin15\degree\\\text{Area of }\triangle ADB&=15.9\text{ cm}^2\text{ (3sf)}\end{aligned}$

Answer: (a) 9.81 cm (b) 15.9 cm $^2$

Q5Swiss Cottage Secondary School2024Prelim P1 Q20●●○ Standard6 marks

Cosine rule, bearings and area of a triangle, Swiss Cottage Secondary School, 2024 Prelim P1 Q20

Swiss Cottage Secondary School 2024 P1 Q20: diagram for the question

The diagram shows the positions of three towns, $A$ , $B$ and $C$ .
$AB$ is 73 km, $BC$ is 62 km and $AC$ is 56 km.
$A$ is due North of $C$ .

Calculate the bearing of $B$ from $A$ . [4]
Calculate the area of triangle $ABC$ . [2]

Show worked solution▾

(a)

$\begin{aligned}62^2&=56^2+73^2-2(56)(73)\cos\angle BAC\\\angle BAC&=55.5845\\\text{Bearing}&=180-55.5845\\&=124.4\end{aligned}$

(b)

$\begin{aligned}&\text{Area of triangle }ABC\\&=\frac12(56)(73)\sin55.5845\\&=1686.2195\\&=1690\text{ km}^2\end{aligned}$

Answer: (a) $124.4^\circ$ (b) $1690$ km $^2$

Q6Greendale Secondary School2025Prelim P1 Q26●●○ Standard4 marks

Angle of depression in a cuboid (3D trigonometry), Greendale Secondary School, 2025 Prelim P1 Q26

The diagram shows a cuboid where $AB=20$ cm and $BC=16$ cm.
$EFGH$ is on a horizontal plane.
It is given that angle $EBF=36^\circ$ .

Greendale Secondary School 2025 P1 Q26: diagram for the question

Calculate

$AE$ , [1]
the angle of depression of $E$ from $C$ . [3]

Show worked solution▾

(a)

$\tan36^\circ=\frac{20}{BF}$

$BF=27.527$

$AE=BF=27.5$

(b)

$AC=\sqrt{16^2+20^2}$

$=\sqrt{656}$

$\tan\angle ACE=\frac{27.527}{\sqrt{656}}$

$\angle ACE=47.0633$

$\text{Angle of depression}=47.1^\circ$

Answer: $\text{(a) }27.5\text{ cm}\quad\text{(b) }47.1^\circ$

Q7Ngee Ann Secondary School2025Prelim P1 Q22●●● Challenging4 marks

Angles of elevation: height of a hill, Ngee Ann Secondary School, 2025 Prelim P1 Q22

Ngee Ann Secondary School 2025 P1 Q22: diagram for the question

Florian wants to find the height of a hill.
He found that the angle of elevation of the peak of the hill, $T$ , from a point $P$ on horizontal ground is $20.8^\circ$ and that from another point $Q$ is $21.7^\circ$ .
$Q$ is 20m closer to the hill than $P$ .

Find the height of the hill in metres. [4]

Show worked solution▾

Ngee Ann Secondary School 2025 P1 Q22: worked solution diagram

Method 1

Let distance $QF$ be $x$ m.

$\tan20.8^\circ=\frac{TF}{20+x}$

$(20+x)\tan20.8^\circ=TF$

$x=\frac{TF-20\tan20.8^\circ}{\tan20.8^\circ}$

$\tan21.7^\circ=\frac{TF}{x}$

$x=\frac{TF}{\tan21.7^\circ}$

So,

$\frac{TF-20\tan20.8^\circ}{\tan20.8^\circ}=\frac{TF}{\tan21.7^\circ}$

$TF\tan21.7^\circ-20\tan20.8^\circ\tan21.7^\circ=TF\tan20.8^\circ$

$TF\tan21.7^\circ-TF\tan20.8^\circ=20\tan20.8^\circ\tan21.7^\circ$

$TF(\tan21.7^\circ-\tan20.8^\circ)=20\tan20.8^\circ\tan21.7^\circ$

$TF=\frac{20\tan20.8^\circ\tan21.7^\circ}{\tan21.7^\circ-\tan20.8^\circ}$

$=167\text{ m (3 s.f.)}$

Method 2:

$\angle PQT=180^\circ-21.7^\circ\quad(\text{adj }\angle\text{s on a st. line})$

$=158.3^\circ$

$\angle PTQ=180^\circ-158.3^\circ-20.8^\circ\quad(\angle\text{ sum of }\triangle)$

$=0.9^\circ$

By Sine Rule,

$\frac{PT}{\sin158.3^\circ}=\frac{20}{\sin0.9^\circ}$

$PT=\frac{20\sin158.3^\circ}{\sin0.9^\circ}$

$=470.79555$

$\sin20.8^\circ=\frac{TF}{470.79555}$

$TF=470.79555\sin20.8^\circ$

$=167\text{ m (3 s.f.)}$

Answer: $167\text{ m}$

Q8North Vista Secondary School2024Prelim P2 Q7●●● Challenging10 marks

Sine rule and greatest angle of depression, North Vista Secondary School, 2024 Prelim P2 Q7

North Vista Secondary School 2024 P2 Q7: diagram for the question

The diagram shows a section of a playground, $ABC$ , bounded by three footpaths $AB$ , $AC$ and $BC$ .
$AB=30$ m, $BC=38$ m and angle $BAC=95^\circ$ .

Find angle $ABC$ . [3]
A vertical pole of 2.5 m is placed at point $A$ .
Calculate the greatest angle of depression of a point along $BC$ from the top of the vertical pole. [3]
Ken runs from $B$ to $C$ at a speed of 4 m/s.
3 seconds after Ken left point $B$ , Ali started to run from $A$ towards $C$ .
Determine the speed that Ali needs to run in order to reach point $C$ at the same time as Ken. [4]

Show worked solution▾

(a)

$\begin{aligned}\frac{\sin\angle ACB}{30}&=\frac{\sin95^\circ}{38}\\\angle ACB&=\sin^{-1}\left(\frac{30\sin95^\circ}{38}\right)\\&=51.856^\circ\\\angle ABC&=180-95-51.856\\&=33.144\\&=33.1^\circ\end{aligned}$

(b)

Let the shortest distance from $A$ to $BC$ be $h$ and greatest angle of depression be $\theta$ .

$\begin{aligned}\sin33.144&=\frac h{30}\\h&=30\times\sin33.144\\&=16.402\ \text{m}\end{aligned}$

$\begin{aligned}\tan\theta&=\frac{2.5}{16.402}\\\theta&=8.6663^\circ\\&=8.7^\circ\ (1\text{ d.p.})\end{aligned}$

(c)

$\begin{aligned}\frac{AC}{\sin33.144}&=\frac{38}{\sin95}\\AC&=\frac{38\sin33.144}{\sin95}\\&=20.855\ \text{m}\end{aligned}$

$\begin{aligned}\text{Time taken by Ken}&=\frac{38}4\\&=9.5\ \text{s}\end{aligned}$

$\begin{aligned}\text{Speed of Ali}&=\frac{20.855}{9.5-3}\\&=3.2085\ \text{m/s}\\&=3.21\ \text{m/s (3 s.f.)}\end{aligned}$

Answer: (a) $33.1^\circ$ (b) $8.7^\circ$ (c) $3.21$ m/s

Q9Anglo-Chinese School (Barker Road)2023Prelim P2 Q8●●● Challenging11 marks

Bearings, cosine rule and greatest angle of elevation, Anglo-Chinese School (Barker Road), 2023 Prelim P2 Q8

The diagram below shows 3 points, $A$ , $B$ and $C$ on a level ground.
The bearing of $A$ from $B$ and $B$ from $C$ are $210^\circ$ and $260^\circ$ respectively.
$AB=75$ m and $BC=110$ m.

Anglo-Chinese School (Barker Road) 2023 P2 Q8: diagram for the question

Show that angle $ABC=130^\circ$ . [2]
Find
1. the distance $AC$ , [3]
2. the bearing of $C$ from $A$ . [3]
A kite was seen flying 55 m vertically above point $B$ . Find the greatest angle of elevation of the kite observed by a person walking in a straight path from $A$ to $C$ . [3]

Show worked solution▾

(a)

$\begin{aligned}\angle a&=360^\circ-260^\circ=100^\circ\\\angle b&=80^\circ\text{(co-interior angles, parallel north lines)}\\\angle ABC&=210^\circ-80^\circ=130^\circ\end{aligned}$

(b)(i)

$\begin{aligned}AC&=\sqrt{75^2+110^2-2(75)(110)\cos130^\circ}\\&=168.318\\&=168\text{m}\end{aligned}$

(b)(ii)

$\begin{aligned}\frac{\sin a}{110}&=\frac{\sin130^\circ}{168.318}\\\angle a&=30.0^\circ\\30^\circ+30^\circ&=060.0^\circ\end{aligned}$

(c)

$\begin{aligned}d&=75\sin30.04^\circ\\&=37.545\text{m}\\\tan\theta&=\frac{55}{37.545}\\\theta&=55.7^\circ\end{aligned}$

Answer: (b)(i) $168$ m (b)(ii) $060.0^\circ$ (c) $55.7^\circ$

Q10Dunman Secondary School2025Prelim P2 Q5●●● Challenging11 marks

Cosine rule and angle of elevation in a 3D solid, Dunman Secondary School, 2025 Prelim P2 Q5

Dunman Secondary School 2025 P2 Q5: diagram for the question

The diagram shows a solid made of a cuboid $ABCDEFGH$ and a triangular prism $EFGHIJ$ , resting on level ground.
$AB=8$ m, $BC=12$ m, $AE=3$ m, $EI=3$ m and $FI=7$ m.

Show that angle $EFI=21.8^\circ$ , correct to 1 decimal place. [3]
Find the volume of the solid, leaving your answer correct to 2 decimal places. [2]
Find the vertical height of $I$ above the ground. [2]
Find the angle of elevation of $J$ from $B$ . [4]