GCSE Sine and cosine rules Practice Questions

State the sine rule.

The sine rule links each side of a triangle to the sine of the angle opposite it: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$. Use it whenever you know a side and the angle opposite it.

Pick out the pair of sides and angles that the question needs.

Side $b$ is opposite angle $B$, and side $a$ is opposite angle $A$. Both of those pairings are known or findable, so the sine rule can start.

State the length of side b.

Side $b$ is $13.5$ cm, correct to $1$ decimal place.

State the sine rule.

The sine rule links each side of a triangle to the sine of the angle opposite it: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$. Use it whenever you know a side and the angle opposite it.

Pick the two fractions that the question needs.

Side $b$ faces angle $B$, and side $a$ faces angle $A$. Both pairs are known or wanted, so these are the two fractions to use.

State the correct calculation.

This is the sine rule rearranged for $b$, with the sine of the angle opposite $b$ on the top.

State the cosine rule.

The cosine rule is $a^2 = b^2 + c^2 - 2bc\cos A$, where $A$ is the angle between the two sides $b$ and $c$. Use it when the sine rule cannot start because no side is paired with its opposite angle.

Rearrange the cosine rule to see what the sign of the cosine tells you.

The numerator of $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ is positive exactly when $b^2 + c^2 > a^2$. So the sign of $\cos A$ — and hence whether the angle is acute or obtuse — is decided by comparing $a^2$ with $b^2 + c^2$.

Find the longest side, because it faces the largest angle.

The largest angle of a triangle is opposite its longest side, $12$ cm. Only that angle can possibly be obtuse, so it is the only one worth testing.

Square the three sides.

The three squares are $49$, $81$ and $144$.

Compare the square of the longest side with the sum of the other two squares.

$130$ is less than $144$, so $12^2 > 7^2 + 9^2$.

State the correct statement.

The angle opposite the $12$ cm side is obtuse.

State the cosine rule.

The cosine rule is $a^2 = b^2 + c^2 - 2bc\cos A$, where $A$ is the angle between the two sides $b$ and $c$. Use it when the sine rule cannot start because no side is paired with its opposite angle.

Match the letters in the rule to the letters in the question.

The angle $A$ is between the two known sides, and $a$ is the side opposite it, which is exactly the arrangement the rule is written for.

Substitute the numbers into the rule.

Putting $b = 13$, $c = 9$ and $A = 112^\circ$ into $a^2 = b^2 + c^2 - 2bc\cos A$ gives this.

Take the square root of both sides, because the rule gives the square.

The cosine rule produces $a^2$. The length itself is the positive square root, so the whole right-hand side must sit under the root sign.

Note the standard mistake with the cosine rule.

The cosine rule SUBTRACTS $2bc\cos A$. It is Pythagoras with a correction term, and the correction is a subtraction; adding it would make every triangle bigger than a right-angled one.

Note the other standard mistake.

The cosine rule gives $a^2$, not $a$. Forgetting the square root leaves an answer many times too big.

Reject the option that drops the 2.

The correction term is $2bc\cos A$. Halving it is not a small slip — it gives a different triangle entirely.

Reject the option that is just Pythagoras.

$\sqrt{b^2 + c^2}$ is Pythagoras, and it is only right when $A = 90^\circ$. Here $A = 112^\circ$, so the correction term cannot be dropped.

Say why the cosine rule is the right rule here.

The sine rule cannot start, because no side is given together with the angle opposite it. That is exactly the situation the cosine rule is for.

State the correct calculation.

This is the cosine rule with $b = 13$, $c = 9$ and $A = 112^\circ$, square-rooted to give a length.

Recognise the configuration.

You are given two sides and an angle that is NOT between them. This is the one configuration that may not pin the triangle down, so the number of triangles has to be tested.

Work out the height from C down to the line AB.

Put $A$ at one end of a line and swing side $b$ up at $63^\circ$ to reach $C$. The shortest distance from $C$ back down to that line is $h = b\sin A = 15.147$ cm.

Compare the swinging side with that height.

Side $a$ starts at $C$ and swings down to meet the line. If it is shorter than $h$ it never gets there; if it is longer it crosses. Here $a = 9$ is shorter than $h = 15.147$.

Compare the swinging side with the fixed side.

If $a$ is at least as long as $b$, only one of the two crossings lies on the correct side of $A$, and there is one triangle. If $a$ is shorter than $b$ but longer than $h$, both crossings are valid.

Apply the three-way test for the ambiguous case.

Here $a = 9 < 15.147 = b\sin A$, so the swinging side is too short to reach the base line at all and no triangle exists.

Say what the two triangles would be, if there are two.

When two triangles exist they share the same $A$, $a$ and $b$, and differ only in angle $B$: one takes the acute inverse sine, the other its obtuse partner $180^\circ - B$.

Ask whether an obtuse angle would also work.

A sine value never picks out one angle on its own: $\sin\theta$ and $\sin(180^\circ - \theta)$ are equal. Whenever the sine rule is used to find an ANGLE you must decide whether the obtuse partner is possible too.

Check the answer against the picture.

The figure shows the arc of radius $a = 9$ cm drawn from $C$: it meets the base line in the number of places the test predicts, so the count is $0$.

Check the labelling convention.

In triangle $ABC$ the side $a$ is opposite angle $A$, side $b$ is opposite angle $B$ and side $c$ is opposite angle $C$. Getting this pairing right is what makes both rules work.

Check the calculator is in degree mode.

Every angle here is in degrees. Test the calculator with $\sin 30^\circ$: it must give $0.5$. In radian mode it gives $-0.988$ and every answer would be wrong.

Note the borderline case.

If $a$ were exactly equal to $b\sin A$ the swinging side would just touch the line, giving a single right-angled triangle. That is the knife edge between two triangles and none.

Note why the other options are impossible.

A circle can cut a straight line in at most two points, so SSA can never give three triangles, and the given lengths fix the shape, so it can never give infinitely many.

Note the rule you would use next.

Whichever count comes out, the way to find angle $B$ is the sine rule, and the count tells you how many of the two inverse sines to keep.

Note that SSS, SAS and AAS never behave like this.

Three sides, or two sides with the angle between them, or two angles with a side, all fix the triangle completely. Only SSA — two sides and an angle NOT between them — can leave a choice.

State how many triangles are possible.

Here $a = 9 < 15.147 = b\sin A$, so the swinging side is too short to reach the base line at all and no triangle exists.