Solution of Triangles Worked Examples (Easy)

What these examples cover

These easy Solution of Triangles examples build the everyday moves that open almost every question in the chapter: using the sine rule to find a missing side, using the cosine rule to find a side when you know two sides and the angle between them, using the cosine rule the other way to find an angle from three sides, and finding the area of a triangle from two sides and their included angle. Each one uses small, clean numbers so you can follow every line while your calculator does only the trigonometry.

Use the set the honest way, cover the solution, attempt the question in full on paper, and only then check line by line against our working. Where your answer differs, find the exact step where the two solutions part company; that single line is usually where the real learning is.

A quick labelled sketch of the triangle keeps each side matched to the angle facing it.

Worked examples

Work through all four. Attempt each fully before you read the matching solution, and notice how the same discipline, pair each side with the angle opposite it, substitute one value at a time, simplify before you round, runs through the sine rule, the cosine rule and the area formula alike.

Q1[3 marks]

In triangle $ABC$ , $\angle A = 30^{\circ}$ , $\angle C = 45^{\circ}$ and side $a = 12$ cm. Find the length of side $c$ .

Show worked solution

Side $a = 12$ sits opposite the known angle $A = 30^{\circ}$ , and we want side $c$ , which sits opposite $C = 45^{\circ}$ . Because we have a complete side–angle pair, the sine rule links them straight away:

\dfrac{a}{\sin A}=\dfrac{c}{\sin C}

Make $c$ the subject, then substitute $a = 12$ , $A = 30^{\circ}$ and $C = 45^{\circ}$ :

c=a\times\dfrac{\sin C}{\sin A}=12\times\dfrac{\sin 45^{\circ}}{\sin 30^{\circ}}

With $\sin 45^{\circ}=0.7071$ and $\sin 30^{\circ}=0.5$ , work the fraction before you multiply:

c=12\times\dfrac{0.7071}{0.5}=12\times1.4142=16.97

Answer

The side $c = 16.97$ cm (2 d.p.). A quick sense check: $c$ faces the larger angle $45^{\circ}$ , so it should be longer than $a = 12$ , and it is.

The exact value is $12\sqrt{2}$ .

Q2[3 marks]

In triangle $ABC$ , $b = 5$ cm, $c = 8$ cm and the included angle $A = 60^{\circ}$ . Find the length of side $a$ .

Show worked solution

Here you know two sides and the angle sitting between them, so the sine rule cannot start, there is no complete side–angle pair yet. The cosine rule is built for exactly this case:

a^{2}=b^{2}+c^{2}-2bc\cos A

Substitute $b = 5$ , $c = 8$ and $A = 60^{\circ}$ , keeping each part clear:

a^{2}=5^{2}+8^{2}-2(5)(8)\cos 60^{\circ}

Since $\cos 60^{\circ}=0.5$ , simplify the numbers one step at a time:

a^{2}=25+64-80(0.5)=89-40=49

Take the positive square root, because a length cannot be negative:

a=\sqrt{49}=7

Answer

The side $a = 7$ cm. Notice the middle term is subtracted only because the angle is acute, for an obtuse angle $\cos A$ would be negative, and that term would add on instead.

Q3[3 marks]

In triangle $ABC$ , the three sides are $a = 3$ cm, $b = 8$ cm and $c = 7$ cm. Find $\angle C$ .

Show worked solution

You are given all three sides and asked for an angle, so use the cosine rule rearranged to make the angle the subject. Angle $C$ faces side $c$ , so $c$ is the side that stands alone:

\cos C=\dfrac{a^{2}+b^{2}-c^{2}}{2ab}

Substitute $a = 3$ , $b = 8$ and $c = 7$ , then simplify the top and bottom separately:

\cos C=\dfrac{3^{2}+8^{2}-7^{2}}{2(3)(8)}=\dfrac{9+64-49}{48}=\dfrac{24}{48}=0.5

Now take the inverse cosine:

C=\cos^{-1}(0.5)=60^{\circ}

Answer

The angle $C = 60^{\circ}$ . Because $\cos C$ came out positive, $C$ is acute, which fits, since $c = 7$ is not the longest side, so it cannot face the largest angle.

Q4[2 marks]

A triangle has two sides of $8$ cm and $9$ cm, with an included angle of $30^{\circ}$ between them. Find the area of the triangle.

Show worked solution

The area formula needs exactly what you are given, two sides and the angle held between them. Call the two sides $a$ and $b$ and the angle between them $C$ :

\text{Area}=\tfrac{1}{2}ab\sin C

Substitute the two sides $8$ and $9$ and the included angle $30^{\circ}$ , with $\sin 30^{\circ}=0.5$ :

\text{Area}=\tfrac{1}{2}(8)(9)\sin 30^{\circ}=36\times0.5=18

Answer

The area is $18$ cm $^{2}$ . The angle must be the one enclosed by the two chosen sides, if you are given a different angle, find an included angle first, or the formula will measure the wrong triangle.

Q5[3 marks]

In triangle $PQR$ , $\angle P = 60^{\circ}$ , $p = 14$ cm and $q = 9$ cm. Find $\angle Q$ .

Show worked solution

Side $p = 14$ sits opposite the known angle $P = 60^{\circ}$ , and side $q = 9$ sits opposite the angle $Q$ you want. With a complete side–angle pair plus one more side, the sine rule can be used again, this time to find an angle:

\dfrac{p}{\sin P}=\dfrac{q}{\sin Q}

Rearrange to make $\sin Q$ the subject, then substitute $p = 14$ , $q = 9$ and $P = 60^{\circ}$ , simplifying the fraction as you go:

\sin Q=\dfrac{q\sin P}{p}=\dfrac{9\sin 60^{\circ}}{14}=\dfrac{9(0.8660)}{14}=\dfrac{7.794}{14}=0.5567

Now take the inverse sine:

Q=\sin^{-1}(0.5567)=33.83^{\circ}

Answer

The angle $Q \approx 33.83^{\circ}$ (2 d.p.). Sense check: side $q = 9$ is shorter than side $p = 14$ , so $Q$ must be smaller than $P = 60^{\circ}$ , and $33.83^{\circ}$ is indeed smaller.

Q6[3 marks]

In triangle $ABC$ , $\angle A = 40^{\circ}$ , $\angle B = 95^{\circ}$ and $c = 10$ cm. Find the length of side $a$ .

Show worked solution

You are given two angles directly, but the side you know ( $c$ ) and the side you want ( $a$ ) sit opposite $C$ and $A$ , and $C$ is not given yet. Find the missing angle first, using the angle sum of a triangle:

\angle C=180^{\circ}-(\angle A+\angle B)=180^{\circ}-(40^{\circ}+95^{\circ})=45^{\circ}

Now side $c = 10$ (opposite the angle $C$ you just found) and side $a$ (opposite $A$ ) form a usable pair, apply the sine rule:

\dfrac{a}{\sin A}=\dfrac{c}{\sin C}

Make $a$ the subject, then substitute $c = 10$ , $A = 40^{\circ}$ and $C = 45^{\circ}$ :

a=c\times\dfrac{\sin A}{\sin C}=10\times\dfrac{\sin 40^{\circ}}{\sin 45^{\circ}}=10\times\dfrac{0.6428}{0.7071}=10\times0.9091=9.09

Answer

Side $a \approx 9.09$ cm (2 d.p.). Sense check: angle $A(40^{\circ})$ is smaller than angle $C(45^{\circ})$ , so side $a$ should be shorter than side $c = 10$ cm, and $9.09$ cm is indeed shorter.

Q7[2 marks]

A triangle has an area of $24$ cm $^{2}$ . Side $a = 8$ cm, and the angle between side $a$ and the unknown side $b$ is $30^{\circ}$ .

Find the length of $b$ .

Show worked solution

This time the area is already known, and a side is missing, rearrange the area formula so the unknown side becomes the subject:

\text{Area}=\tfrac{1}{2}ab\sin C \quad\Rightarrow\quad b=\dfrac{2\times\text{Area}}{a\sin C}

Substitute Area $=24$ , $a=8$ , and the included angle $C=30^{\circ}$ , with $\sin 30^{\circ}=0.5$ :

b=\dfrac{2(24)}{8\sin 30^{\circ}}=\dfrac{48}{8(0.5)}=\dfrac{48}{4}=12

Answer

Side $b = 12$ cm. Sense check: substitute back, Area $=\tfrac{1}{2}(8)(12)\sin 30^{\circ}=\tfrac{1}{2}(96)(0.5)=24$ cm $^{2}$ , which matches the given area exactly.

Q8[3 marks]

In triangle $ABC$ , $AB = 10$ cm, $BC = 8$ cm and $\angle B = 50^{\circ}$ . Find the perimeter of the triangle.

Show worked solution

The perimeter needs all three sides, but only two are given, first find the third side $AC$ using the cosine rule, since you know two sides and their included angle $B$ :

AC^{2}=AB^{2}+BC^{2}-2(AB)(BC)\cos B

Substitute $AB=10$ , $BC=8$ and $B=50^{\circ}$ , with $\cos 50^{\circ}=0.6428$ :

AC^{2}=10^{2}+8^{2}-2(10)(8)(0.6428)=164-102.85=61.15

Take the positive square root:

AC=\sqrt{61.15}=7.82

Now add all three sides to get the perimeter:

\text{Perimeter}=AB+BC+AC=10+8+7.82=25.82

Answer

The perimeter $\approx 25.82$ cm (2 d.p.). Sense check: $AC (7.82$ cm $)$ is shorter than both given sides, which fits a modest included angle of $50^{\circ}$ .

Notice how the choice of rule follows only from what you are given. A complete side–angle pair points to the sine rule; two sides with the angle between them, or all three sides, points to the cosine rule; and two sides plus their included angle gives the area at once.

Read exactly what the question hands you, match it to the right tool, and the arithmetic stays short and reliable.

Key method points

These four examples rehearse the tools that open almost every Solution of Triangles question in Add Math. Keep the following points in mind as you practise more.

Pair each side with the angle facing it: side $a$ is opposite $\angle A$ , and so on. This decides which rule you can start with.
Use the sine rule $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$ when you already have one complete side–angle pair.
Use the cosine rule $a^{2}=b^{2}+c^{2}-2bc\cos A$ for two sides and the included angle (to find the third side), or rearranged as $\cos A=\dfrac{b^{2}+c^{2}-a^{2}}{2bc}$ for three sides (to find an angle).
A negative cosine means the angle is obtuse; a positive cosine means it is acute, a fast check against your sketch.
The area of a triangle is $\tfrac{1}{2}ab\sin C$ , where $C$ is the angle enclosed by the two sides $a$ and $b$ .
Take only the positive square root for a length, and keep every substitution line, with analytic marking a clear method line still earns method marks even if the final digit slips.

How a teacher helps

When a student drops a mark on questions like these, it is nearly always a small, fixable habit, reaching for the sine rule when no side–angle pair is complete, or mismatching a side with the wrong angle. In a one-to-one lesson our teacher watches the exact line where the slip happens and corrects it on the spot, before it settles into a routine.

Because our teachers are experienced, you work with someone who explains why each rule fits its situation, not just how to press the keys. Lessons are taught in English, while SPM papers are set in both Malay and English, so the notation reads the same to you either way.

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Frequently asked questions

How do I decide between the sine rule and the cosine rule?

Look at what you are given. If you already have a matching side and its opposite angle, start with the sine rule.

If you have two sides and the angle between them, or all three sides, use the cosine rule. The sine rule needs a complete side–angle pair to get going; the cosine rule does not.

What does "included angle" mean?

It is the angle sitting between the two sides you are using, the corner where those two sides meet. The cosine rule for a side and the area formula $\tfrac{1}{2}ab\sin C$ both need the angle to be the included one, not one of the other two.

How many decimal places should I keep?

Carry a few extra figures through the working and round only at the end, usually to two decimal places for a length or an angle, or as the question states. Rounding a trigonometric value too early can shift the final digit and cost accuracy.

Does it matter which letters I use for the sides and angles?

The labels are just names, but the pairing is fixed: the side and the angle that share a letter must sit opposite each other. Keep $a$ opposite $A$ , $b$ opposite $B$ and $c$ opposite $C$ , and every formula on this page works exactly as written.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

Solution of Triangles, Worked Examples (easy)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

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