Hard GCSE Sine and cosine rules Questions

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.

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

Work out the third angle.

Side $c$ is opposite angle $C$, which is not given, so find it first from the angle sum: $180 - 27 - 118 = 35$.

Rearrange the sine rule to make the unknown side the subject.

Multiplying both sides by $\sin C$ gives $c = \frac{a \sin C}{\sin A}$. The sine of the angle OPPOSITE the side you want goes on the top.

Substitute the numbers into the rearranged rule.

The known side is $21$ cm with its opposite angle $27^\circ$, and the angle opposite $c$ is $35^\circ$.

Work out the two sines.

$\sin 35^\circ = 0.5736$ and $\sin 27^\circ = 0.4540$, both from the calculator in degree mode.

Divide, then round to the accuracy asked for.

The calculator gives $26.53162\ldots$, and rounding to $2$ decimal places gives $26.53$.

Say why the sine rule is the right rule here.

The sine rule needs a matching pair: a side together with the angle opposite it. That pair is given here, so the sine rule starts straight away and the cosine rule is not needed.

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.

Estimate the answer before working it out.

The bigger angle faces the longer side, and here $C$ is bigger than $A$. So the answer must lie between about $26.0$ and $27.0$; anything outside that range is wrong.

Check the longest side is opposite the largest angle.

The longest side $b$ sits opposite the largest angle $B$, and the shortest side $a$ sits opposite the smallest angle $A$. If that failed, the answer would be wrong.

Check the three angles add up to 180 degrees.

The angles of the solved triangle are about $27.0^\circ$, $118.0^\circ$ and $35.0^\circ$, and they add to $180^\circ$, so the triangle is consistent.

Check the triangle inequality.

The two shorter sides add to about $47.5$, which is more than the longest side $40.8$, so a triangle with these sides really exists.

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.

State the length of side c.

Side $c$ is $26.53$ cm, correct to $2$ decimal places.

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, $17$ cm. Only that angle can possibly be obtuse, so it is the only one worth testing.

Square the three sides.

The three squares are $81$, $121$ and $289$.

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

$202$ is less than $289$, so $17^2 > 9^2 + 11^2$.

Read off whether the largest angle is acute or obtuse.

The cosine of the largest angle is negative, so that angle is obtuse. The comparison of squares said the same thing without a calculator.

Reject the statements about the two shorter sides.

A shorter side can never have a square bigger than the sum of the other two squares, so no angle except the largest can be obtuse. Both of those options are false whatever the numbers.

Note why the cosine rule has no ambiguous case.

Unlike the sine, the cosine has a different SIGN for acute and obtuse angles, so a cosine value picks out exactly one angle between $0^\circ$ and $180^\circ$. Finding an angle with the cosine rule can never be ambiguous.

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 triangle inequality.

The two shorter sides add to about $20.0$, which is more than the longest side $17.0$, so a triangle with these sides really exists.

Check the longest side is opposite the largest angle.

The longest side $c$ sits opposite the largest angle $C$, and the shortest side $a$ sits opposite the smallest angle $A$. If that failed, the answer would be wrong.

Check the three angles add up to 180 degrees.

The angles of the solved triangle are about $28.4^\circ$, $35.5^\circ$ and $116.1^\circ$, and they add to $180^\circ$, so the triangle is consistent.

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.

Sketch the solved triangle to check it looks right.

Drawn to scale, the triangle has sides of about $9.0$, $11.0$ and $17.0$, and the picture matches the answer.

State the correct statement.

The angle opposite the $17$ cm side is obtuse.

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.

Plan the work: find angle B first, then side c.

Each of the two values of $B$ gives its own third angle $C$, and each $C$ gives its own side $c$. So there are two possible lengths, not one.

Check that the data really does give two triangles.

The perpendicular distance from $C$ to the line $AB$ is $b\sin A = 10.420$ cm. The side $a = 12$ cm is longer than that, so it reaches the line, but it is shorter than $b = 15$ cm, so it reaches it TWICE. That is the ambiguous case: two triangles.

Rearrange the sine rule for the unknown angle and substitute.

Side $b$ is opposite angle $B$ and side $a$ is opposite angle $A$, so $\frac{\sin B}{b} = \frac{\sin A}{a}$ gives $\sin B = \frac{b \sin A}{a}$.

Work out the sine of the unknown angle.

$\sin B = \frac{15 \times 0.6947}{12} = 0.8683$, which is less than $1$, so an angle $B$ does exist.

Take the inverse sine to get the acute solution.

The calculator returns only the acute angle, about $60.3^\circ$. This is the first of the two answers, not the only one.

Find the obtuse angle that has the same sine.

$\sin\theta = \sin(180^\circ - \theta)$, so $119.7^\circ$ has the same sine as $60.3^\circ$ and is a second candidate for $B$.

Check that the obtuse angle really leaves a triangle.

Adding the obtuse candidate to $A = 44^\circ$ still leaves $16.3^\circ$ for angle $C$, which is positive, so the obtuse triangle genuinely exists. This is the test that decides the ambiguous case, and it is the step candidates most often skip.

Work out the third angle of each triangle.

Subtracting each value of $B$ from $180^\circ - 44^\circ$ gives the two third angles, about $75.7^\circ$ and $16.3^\circ$.

Use the sine rule once more for each possible side c.

With the acute $B$ the triangle is long and flat and $c = 16.7$ cm; with the obtuse $B$ it is short and steep and $c = 4.8$ cm.

Reject the distractors built from part-answers.

$10.8$ is only the projection $b\cos A$ of side $b$ onto $AB$, and $6.0$ is only half the gap between the two answers. Each is a stepping stone, not a possible value of $c$.

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.

Write down both complete triangles.

The acute solution gives the long, flat triangle with $c \approx 16.7$ cm; the obtuse solution gives the short, steep one with $c \approx 4.8$ cm. Both fit the given $A$, $a$ and $b$ exactly.

Check the angle sum of the acute triangle.

The acute triangle has angles about $44^\circ$, $60.3^\circ$ and $75.7^\circ$, which add to $180^\circ$.

State both possible lengths of side c.

Both lengths give a genuine triangle, so both must be given.

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.

Draw the triangle and see which rule fits.

The two legs $PQ$ and $QR$ meet at $Q$, and the angle at $Q$ is the angle between them. Two sides and the angle between them is exactly the cosine rule.

Substitute the two sides and the angle between them.

The two known sides are $19$ and $23$, and $133^\circ$ is the angle BETWEEN them, which is exactly what the cosine rule needs.

Square the two known sides and add them.

$19^2 = 361$ and $23^2 = 529$, and together they make $890$. So far this is just Pythagoras.

Work out the correction term.

$\cos 133^\circ = -0.6820$, so the correction term is $-596.067$. It is negative, because the angle is obtuse, so subtracting it makes the side LONGER than Pythagoras would give.

Subtract to get the square of the unknown side.

$PR^2 = 1486.067$. This is the SQUARE of the side, not the side.

Take the positive square root and round.

A length cannot be negative, so take the positive square root: $38.5495\ldots$, which is $38.5$ to $1$ decimal place.

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.

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.

Estimate the answer before working it out.

The angle between the two sides is obtuse, so the third side must be LONGER than it would be in a right-angled triangle. So the answer must lie between about $38.0$ and $39.0$; anything outside that range is wrong.

Check the answer by putting it back into the cosine rule.

Squaring the answer gives about $1486.067$, which is what the right-hand side of the cosine rule came to, so the answer fits.

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.

Do not round anything until the very end.

Rounding a value part-way through and then using the rounded value makes the final answer drift. Keep the full calculator display and round only the last line.

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.

State the distance PR.

The distance $PR$ is $38.5$ km, correct to $1$ decimal place.