Solution of Triangles Practice

Q: What is the ambiguous case?

When the sine rule gives sin⁡B\sin BsinB, the angle BBB may be acute or obtuse, because sin⁡B=sin⁡(180∘−B)\sin B=\sin(180^{\circ}-B)sinB=sin(180∘−B). If both values leave a positive third angle, two triangles are possible and you must give both.

How to use these practice questions

The six questions below rise in difficulty across the whole Solution of Triangles chapter, from a single cosine-rule side up to the ambiguous case with two possible triangles. Give yourself roughly five to eight minutes per question and work on paper first, writing every line the way you would in the real exam, a method line stating the rule you are using, a clear substitution, then the final answer rounded sensibly.

Set your calculator to degrees. Resist the urge to peek; only once you have committed to a full answer should you open the solution and mark yourself line by line.

When your working differs from ours, stop at the exact line where the two part company; that single step is almost always where the mark was lost. Because Add Math is marked analytically, a correct rule and a correct substitution still earn credit even when the final arithmetic slips, so always write the substitution in full before you reach for the calculator.

Treat this page as a rehearsal, not a test, the point is to find weak steps now, while there is still time to fix them.

Six practice questions

Q1[3 marks]

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

Show worked solution

Two sides and the angle between them point straight to the cosine rule, $a^{2}=b^{2}+c^{2}-2bc\cos A$ . Substitute the values, keeping $\cos 60^{\circ}=\tfrac{1}{2}$ :

a^{2}=5^{2}+8^{2}-2(5)(8)\cos 60^{\circ}=25+64-80\left(\tfrac{1}{2}\right)

Simplify, then take the positive square root because a length cannot be negative:

a^{2}=89-40=49\;\Rightarrow\; a=\sqrt{49}=7

Answer

$a=7\ \text{cm}$ . The included angle $60^{\circ}$ is chosen because it sits between the two known sides, that is the exact condition for the cosine rule to apply directly.

Q2[3 marks]

In triangle $ABC$ , $A=45^{\circ}$ , $B=30^{\circ}$ and side $a=10\ \text{cm}$ . Find the length of side $b$ .

Show worked solution

You have two angles and the side opposite one of them, so use the sine rule, $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}$ . Rearrange to make $b$ the subject:

b=\frac{a\sin B}{\sin A}=\frac{10\sin 30^{\circ}}{\sin 45^{\circ}}

Substitute the exact values $\sin 30^{\circ}=\tfrac{1}{2}$ and $\sin 45^{\circ}=\tfrac{\sqrt{2}}{2}$ :

b=\frac{10\left(\tfrac{1}{2}\right)}{\tfrac{\sqrt{2}}{2}}=\frac{5}{\tfrac{\sqrt{2}}{2}}=\frac{10}{\sqrt{2}}=5\sqrt{2}\approx 7.07

Answer

$b=5\sqrt{2}\approx 7.07\ \text{cm}$ . Sanity check: $B=30^{\circ}$ is smaller than $A=45^{\circ}$ , so its opposite side $b$ should be shorter than $a=10$ , and $7.07<10$ , as expected.

Q3[4 marks]

A triangle has sides $a=7\ \text{cm}$ , $b=5\ \text{cm}$ and $c=3\ \text{cm}$ . Find the largest angle of the triangle.

Show worked solution

The largest angle always faces the longest side, so it is angle $A$ , opposite $a=7$ . With three sides known, use the cosine rule rearranged for the angle:

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

Substitute $a=7$ , $b=5$ , $c=3$ , working the numerator and denominator separately:

\cos A=\frac{5^{2}+3^{2}-7^{2}}{2(5)(3)}=\frac{25+9-49}{30}=\frac{-15}{30}=-\frac{1}{2}

A negative cosine means an obtuse angle. Take the inverse cosine:

A=\cos^{-1}\!\left(-\tfrac{1}{2}\right)=120^{\circ}

Answer

The largest angle is $A=120^{\circ}$ . The negative value of $\cos A$ is the signal that the angle is obtuse; if you had found a positive cosine here, that would warn you of a slip in the substitution.

Q4[4 marks]

In triangle $ABC$ , $a=6\ \text{cm}$ , $b=10\ \text{cm}$ and the included angle $C=30^{\circ}$ . Find the area of the triangle.

Show worked solution

The area of a triangle from two sides and the included angle is $\text{Area}=\tfrac{1}{2}ab\sin C$ . The angle $C$ sits between sides $a$ and $b$ , so the rule applies directly:

\text{Area}=\frac{1}{2}(6)(10)\sin 30^{\circ}

Use $\sin 30^{\circ}=\tfrac{1}{2}$ and simplify:

\text{Area}=\frac{1}{2}(60)\left(\tfrac{1}{2}\right)=30\times\tfrac{1}{2}=15

Answer

The area is $15\ \text{cm}^{2}$ . The two sides in the formula must be the ones enclosing the given angle; if the angle offered is not between the two sides, find a different pair or another angle first.

Q5[5 marks]

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

Show worked solution

(a) Two sides and the angle between them call for the cosine rule, $c^{2}=a^{2}+b^{2}-2ab\cos C$ . Substitute with $\cos 60^{\circ}=\tfrac{1}{2}$ :

c^{2}=8^{2}+5^{2}-2(8)(5)\cos 60^{\circ}=64+25-80\left(\tfrac{1}{2}\right)

c^{2}=89-40=49\;\Rightarrow\; c=\sqrt{49}=7

(b) With the same two sides and included angle, the area is $\tfrac{1}{2}ab\sin C$ . Use $\sin 60^{\circ}=\tfrac{\sqrt{3}}{2}$ :

\text{Area}=\frac{1}{2}(8)(5)\sin 60^{\circ}=20\times\frac{\sqrt{3}}{2}=10\sqrt{3}\approx 17.32

Answer

$c=7\ \text{cm}$ and the area is $10\sqrt{3}\approx 17.32\ \text{cm}^{2}$ . Notice the same two sides and the same included angle drive both parts, the cosine rule for the length, the sine formula for the area.

Q6[5 marks]

In triangle $ABC$ , $a=8\ \text{cm}$ , $b=10\ \text{cm}$ and $A=40^{\circ}$ . Find the two possible values of angle $B$ , giving each to two decimal places.

Show worked solution

You have two sides and an angle opposite one of them, so use the sine rule and solve for $\sin B$ . Rearrange $\dfrac{\sin B}{b}=\dfrac{\sin A}{a}$ :

\sin B=\frac{b\sin A}{a}=\frac{10\sin 40^{\circ}}{8}

Evaluate with the calculator in degrees, using $\sin 40^{\circ}\approx 0.6428$ :

\sin B=\frac{10(0.6428)}{8}=\frac{6.428}{8}\approx 0.8035

Because $b>a$ , the angle $B$ can be acute or obtuse, so take both the calculator value and its supplement $180^{\circ}-B$ :

B=\sin^{-1}(0.8035)\approx 53.46^{\circ}\quad\text{or}\quad B=180^{\circ}-53.46^{\circ}=126.54^{\circ}

Answer

$B\approx 53.46^{\circ}$ or $B\approx 126.54^{\circ}$ . Both are valid: with $A=40^{\circ}$ , the remaining angle $C$ would be $86.54^{\circ}$ or $13.46^{\circ}$ , and both are positive, so two genuine triangles exist.

This is the ambiguous case, never forget the second angle.

How to mark yourself like an examiner

Add Math is marked analytically, which means marks are attached to steps, not only to the final number. When you check your own script, award yourself credit the way a marker would: look for the correct rule chosen, the right values put into it, and a clean final answer stated to a sensible accuracy with units.

Method mark: did you choose the correct rule, cosine rule for two sides and the included angle, or three sides; sine rule for an angle-side pair; $\tfrac{1}{2}ab\sin C$ for area?
Substitution mark: are the correct sides and angles placed correctly, with the calculator set to degrees?
Answer mark: is the final value stated clearly, rounded sensibly, and does a quick sanity check on side and angle sizes agree?
For the ambiguous case, you only earn full marks by giving both the acute and the obtuse value of the angle.
If your final number is wrong but the rule and substitution lines are right, give yourself those marks, that is exactly what a real marker does.

How a teacher helps

Marking yourself is powerful, but it is hard to see your own blind spots. In a one-to-one lesson our teacher watches the exact line where a mark slips away, the wrong rule chosen for the given information, a calculator left in radians, or only one angle given in the ambiguous case, and corrects the habit on the spot.

Because our teachers are experienced, you work with someone who explains the why behind each step. Lessons are taught in English, while SPM papers are set in both Malay and English, so we make sure the notation reads the same to you either way.

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Frequently asked questions

How long should each of these questions take me?

Aim for roughly five to eight minutes each, rising with the mark value. If a question takes far longer, note it and bring it to a lesson, the time it steals in the exam is often the real problem, not the topic itself.

When do I use the sine rule and when the cosine rule?

Use the cosine rule when you have two sides and the angle between them, or all three sides. Use the sine rule when you have a matching pair, a side and its opposite angle, plus one more piece of the same kind.

What is the ambiguous case?

When the sine rule gives $\sin B$ , the angle $B$ may be acute or obtuse, because $\sin B=\sin(180^{\circ}-B)$ . If both values leave a positive third angle, two triangles are possible and you must give both.

Do I lose all the marks if my final answer is wrong?

No. Because marking is analytic, a correct rule line and a correct substitution still earn marks even if the arithmetic slips at the end.

That is why you should always show full working before using the calculator.

Source:SRC-DSKP-EN

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

Solution of Triangles, Practice Questions

How to use these practice questions

Six practice questions

How to mark yourself like an examiner

How a teacher helps

Frequently asked questions

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