The cosine rule for HSC Maths Standard 2

Syllabus dot point

“Use the cosine rule to find a side given two sides and the included angle, or an angle given three sides”

HSCMaths Standard 2Year 12: Measurement12 min readUpdated 20 Jun 2026

Quick answer

For SAS use $c^2 = a^2 + b^2 - 2ab\cos C$ to find the third side; for SSS use $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ to find an angle (with $a$ opposite the wanted angle); watch the sign for obtuse angles, take the square root for a side, and remember this is Pythagoras generalised to any triangle.

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What this dot point is asking
The answer
Exam-style questions
Practice questions

What this dot point is asking

NESA wants you to apply the cosine rule in the two situations the sine rule cannot handle. The first is finding the third side when you know two sides and the angle between them (SAS). The second is finding an angle when you know all three sides (SSS). The rule is printed on the NESA reference sheet, so you do not need to memorise it. The marks come from choosing it correctly, substituting accurately, and rounding sensibly.

The deciding question is always the same: can I pair a side with the angle opposite it? In SAS the given angle sits between the two given sides, so it has no opposite side among the ones you know. In SSS there are no angles at all. Either way there is no usable pair, so the sine rule cannot start and the cosine rule takes over.

The answer

The cosine rule for a side (SAS)

For any triangle $ABC$ with sides $a$ , $b$ , $c$ opposite angles $A$ , $B$ , $C$ :

c^2 = a^2 + b^2 - 2 a b \cos C.

Here $c$ is the side opposite the known angle $C$ , and $a$ , $b$ are the two sides that enclose $C$ . By symmetry the same pattern rotates around the triangle:

a^2 = b^2 + c^2 - 2 b c \cos A, \qquad b^2 = a^2 + c^2 - 2 a c \cos B.

The pattern in words: the unknown side squared equals the sum of the squares of the other two sides, minus twice their product times the cosine of the angle between them.

The cosine rule for an angle (SSS)

Rearranging the same equation to make the angle the subject:

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

The angle $A$ is opposite the side $a$ . To find a specific angle, put its opposite side as $a$ in this formula. A neat feature: the sign of the top tells you the angle's type. If $b^2 + c^2 - a^2$ is negative, $\cos A$ is negative and the angle is obtuse; if it is zero the angle is exactly $90\degree$ ; if positive, acute.

Key fact

Use $c^2 = a^2 + b^2 - 2ab\cos C$ to find a side from SAS, and $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ to find an angle from SSS. In both, the angle and the side $c$ (or $a$ ) sit opposite each other, and the two sides multiplied together are the ones that enclose that angle. When $C = 90\degree$ , $\cos C = 0$ and the rule collapses to $c^2 = a^2 + b^2$ , so the cosine rule is just Pythagoras extended to any triangle.

When to use it (and when not to)

SAS (two sides and the included angle): use the side form to get the third side.
SSS (three sides, no angle): use the angle form to get one angle. After that you can use the cosine rule again, or switch to the (quicker) sine rule for the remaining angles.

For AAS / ASA or SSA, there is a complete side-angle pair, so use the sine rule instead.

Identifying the included angle

The included angle is the one between the two given sides. In worded problems it shows up as "the angle at $B$ between $AB$ and $BC$ ", "an angle of $60\degree$ between the two roads", or as the bend in a journey. If the angle you are given is not enclosed by your two sides, you do not have a clean SAS setup, so re-read or redraw before substituting.

The cosine rule, stage by stage (SAS)

Two roads leave a town at an angle of $80\degree$ . A car drives $30$ km along one and $40$ km along the other. How far apart are the two endpoints?

Stage 1, label the SAS triangle. Put the town at vertex $C$ . The two roads are the sides $a = 30$ and $b = 40$ , and the angle between them is $C = 80\degree$ . The distance we want, $c$ , is the side opposite $C$ .

Stage 2, confirm the angle is included. The $80\degree$ sits between the two known sides, so this is genuinely SAS and the cosine rule applies directly. If the angle had been at one of the far ends instead, this would not be SAS.

Stage 3, substitute and solve. $c^2 = 30^2 + 40^2 - 2(30)(40)\cos 80\degree = 900 + 1600 - 2400(0.1736) = 2083.3$ , so $c \approx \sqrt{2083.3} \approx 45.64$ km. Sense-check: $c$ should be shorter than the two roads added end to end ( $70$ km) but longer than their difference ( $10$ km), and $45.64$ sits comfortably between.

The link to Pythagoras

Suppose the included angle is a right angle. Then $\cos 90\degree = 0$ , the term $-2ab\cos C$ becomes zero, and the rule reduces to $c^2 = a^2 + b^2$ . So Pythagoras is just the special case of the cosine rule for right-angled triangles. Put the other way round, the cosine rule is Pythagoras plus a correction term, $-2ab\cos C$ , that adjusts for the angle not being $90\degree$ . When $C$ is acute the correction is negative, so the opposite side is shorter than Pythagoras would predict. When $C$ is obtuse, $\cos C < 0$ , so the correction is positive and the opposite side is longer.

Watch the sign when the angle is obtuse

For an obtuse included angle, $\cos C$ is negative, so $-2ab\cos C$ becomes a positive number you add. Forgetting this is a frequent error. For example with $C = 110\degree$ , $\cos 110\degree \approx -0.342$ , and $-2ab\cos C = +0.684 ab$ . The opposite side comes out longer than either enclosing side, which matches the geometry of a wide-open triangle.

How exam questions ask about the cosine rule

"Find the length of ... / how far apart ..." with two sides and the angle between them. SAS: use the side form.
"Find the size of the largest / smallest angle" with three sides given. SSS: use the angle form. The largest angle faces the longest side and the smallest faces the shortest, so you can target the one asked for directly.
"A ship sails ... then changes course ..." Convert the change of bearing into the interior angle at the bend (often the supplement of the turn), then it is SAS.
"Show that the angle is obtuse / acute." Use the angle form and read the sign of $b^2 + c^2 - a^2$ : negative means obtuse, zero means right, positive means acute.
"Hence find the area / the remaining angles." A multi-part question: use the cosine rule first, then feed the result into the area formula or the sine rule.

Worked example

How far apart are two roads' endpoints (SAS)

Two roads leave a town at $80\degree$ . A car drives $30$ km along one and $40$ km along the other. How far apart are the endpoints?

Set up: This is SAS with $a = 30$ , $b = 40$ and included angle $C = 80\degree$ , so the cosine rule gives the third side $c$ directly.

Substitute into the side form:

c^2 = a^2 + b^2 - 2ab\cos C = 30^2 + 40^2 - 2 \times 30 \times 40 \times \cos 80\degree

Using $\cos 80\degree \approx 0.1736$ :

c^2 = 900 + 1600 - 2400 \times 0.1736 = 2500 - 416.7 = 2083.3

Take the square root:

c \approx \sqrt{2083.3} \approx 45.64 \text{ km}

Answer: the endpoints are about $45.64$ km apart.

Angle opposite the 7 cm side (SSS)

A triangle has sides $5$ cm, $7$ cm and $10$ cm. Find the angle opposite the $7$ cm side.

Substitute into the angle form: Put the wanted angle's opposite side as $a = 7$ , and the two enclosing sides as $b = 5$ , $c = 10$ . Use the angle form:

\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{5^2 + 10^2 - 7^2}{2 \times 5 \times 10} = \frac{25 + 100 - 49}{100} = \frac{76}{100} = 0.76

Invert to find the angle:

A = \cos^{-1}(0.76) \approx 40.5\degree

The top was positive, so we correctly get an acute angle.

Answer: the angle opposite the $7$ cm side is about $40.5\degree$ .

Distance from start after a change of course

A yacht leaves Sydney Heads on a bearing of $130\degree$ for $25$ km, then changes course to a bearing of $200\degree$ for $40$ km. How far is it from the start?

Find the interior angle: The course turns by $200\degree - 130\degree = 70\degree$ . The interior angle of the triangle at the bend is the supplement, $180\degree - 70\degree = 110\degree$ , because the "straight ahead" line and the new leg make a $70\degree$ angle on the outside.

Apply the cosine rule with an obtuse angle: The two legs $25$ km and $40$ km enclose the $110\degree$ angle. Since $\cos 110\degree \approx -0.342$ , the last term is added:

c^2 = 25^2 + 40^2 - 2 \times 25 \times 40 \times \cos 110\degree = 625 + 1600 - 2000(-0.342) = 2225 + 684 = 2909

c \approx \sqrt{2909} \approx 53.94 \text{ km}

Answer: the yacht is about $53.94$ km from its starting point.

Exam technique

State the rule, substitute, then evaluate; markers give method marks for a correct statement and substitution even if the arithmetic slips. For a side, remember the final square root, $c^2$ is not $c$ . For an angle, put the opposite side as $a$ and read the sign of the numerator to predict acute versus obtuse. Keep the calculator in degrees and carry at least four decimal places through to the last line. For an obtuse included angle the $\cos$ is negative, so the formula effectively adds, double-check that sign. And before reaching for any rule on three sides, glance for a Pythagorean triple (like $3$ - $4$ - $5$ or $5$ - $12$ - $13$ ): if the sides fit, the triangle is right-angled and you can skip straight to the simpler tools.

Common traps

Using the wrong included angle: Only the angle between the two given sides works in the side form. Sketch and label.
Forgetting the square root: After finding $c^2$ , take the positive square root to get $c$ .
Sign error with obtuse angles: $\cos$ of an angle between $90\degree$ and $180\degree$ is negative, so $-2ab\cos C$ adds to the total.
Trying the sine rule on SAS or SSS: There is no complete side-angle pair, so it cannot start. Use the cosine rule.
Rounding mid-computation: Carry at least four decimals through; the final answer loses accuracy quickly otherwise.

Note

Picture a triangle that does not have a square corner. Pythagoras and the basic trig ratios only work on triangles with a square corner, so for these slanted triangles you need a stronger tool, and the cosine rule is it. You reach for it in two cases. In the first, you know the lengths of two sides and the size of the angle squeezed between them, and you want the length of the far side facing that angle. In the second, you know all three side lengths and you want the size of one of the angles. The same rule covers both jobs, just written two ways, and the angle you care about always sits opposite the side you are pairing it with.

Exam-style questions

Questions in the style of NESA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

2022 HSC-style3 marks

In triangle

ABC

a = 7

cm,

b = 9

cm and

C = 65\degree

. Find side

c

correct to one decimal place.

Show worked answer →

Use $c^2 = a^2 + b^2 - 2 a b \cos C$ .

$c^2 = 49 + 81 - 2 \times 7 \times 9 \times \cos 65\degree$ .

$\cos 65\degree \approx 0.4226$ .

$c^2 = 130 - 126 \times 0.4226 = 130 - 53.25 = 76.75$ .

$c = \sqrt{76.75} \approx 8.76$ .

Rounded to one decimal place: $c \approx 8.8$ cm.

Markers reward the cosine rule stated, correct substitution, intermediate computation kept to at least four decimal places, and final answer at the requested precision.

2023 HSC-style4 marks

A triangular paddock has sides

80

60

m and

50

m. Find the largest angle of the paddock correct to the nearest degree.

Show worked answer →

The largest angle is opposite the longest side ( $80$ m). Use the cosine rule rearranged:

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

Let $a = 80$ , $b = 60$ , $c = 50$ .

$\cos A = \frac{60^2 + 50^2 - 80^2}{2 \times 60 \times 50} = \frac{3600 + 2500 - 6400}{6000} = \frac{-300}{6000} = -0.05$ .

$A = \cos^{-1}(-0.05) \approx 92.87\degree$ , round to $93\degree$ .

Markers reward identifying the largest angle as opposite the longest side, the rearranged cosine rule, and the angle rounded as requested. Half a mark if you find $A$ but do not justify which side it sits opposite.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation1 mark

In triangle

ABC

, the sides

a = 6

cm and

b = 10

cm meet at an included angle

C = 50^\circ

. Find the length of the third side

c

, correct to one decimal place.

Show worked solution →

State the rule (SAS). The angle $C$ lies between the two known sides, so use the side form:

c^2 = a^2 + b^2 - 2 a b \cos C

Substitute the given values. With $a = 6$ , $b = 10$ and $\cos 50^\circ \approx 0.6428$ :

c^2 = 6^2 + 10^2 - 2 \times 6 \times 10 \times \cos 50^\circ = 36 + 100 - 120 \times 0.6428 = 58.87

Take the square root.

c \approx \sqrt{58.87} \approx 7.7 \text{ cm}

Answer: $c \approx 7.7$ cm.

foundation2 marks

A triangle has sides of length

4

cm,

5

cm and

6

cm. Find the size of the angle opposite the

4

cm side, correct to the nearest degree.

Show worked solution →

State the rule (SSS). Three sides and no angle means use the angle form, with the wanted angle's opposite side as $a = 4$ and the enclosing sides as $b = 5$ and $c = 6$ :

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

Substitute the given values.

\cos A = \frac{5^2 + 6^2 - 4^2}{2 \times 5 \times 6} = \frac{25 + 36 - 16}{60} = \frac{45}{60} = 0.7500

Invert to find the angle.

A = \cos^{-1}(0.7500) \approx 41^\circ

Answer: the angle opposite the $4$ cm side is about $41^\circ$ .

foundation2 marks

In triangle

ABC

, the side

a = 8

cm and the side

b = 11

cm meet at an included angle

C = 40\degree

. Find the length of the third side

c

, correct to two decimal places.

Show worked solution →

State the rule (SAS). The angle $C$ sits between the two known sides, so use the side form of the cosine rule:

c^2 = a^2 + b^2 - 2 a b \cos C

Substitute the given values. With $a = 8$ , $b = 11$ and $C = 40\degree$ , and using $\cos 40\degree \approx 0.7660$ :

c^2 = 8^2 + 11^2 - 2 \times 8 \times 11 \times \cos 40\degree = 64 + 121 - 176 \times 0.7660 = 50.18

Take the square root.

c \approx \sqrt{50.18} \approx 7.08 \text{ cm}

Check. The included angle is acute, so $c$ should be shorter than the Pythagoras value $\sqrt{8^2 + 11^2} \approx 13.60$ cm, and it should sit between the difference of the sides ( $3$ cm) and their sum ( $19$ cm). Since $7.08$ cm satisfies both, the answer is reasonable.

foundation3 marks

A triangular garden bed has sides of length

5

6

m and

9

m. Find the size of the largest angle, correct to the nearest degree.

Show worked solution →

Identify the largest angle. The largest angle lies opposite the longest side, which is the $9$ m side. Put that side as $a = 9$ and the two enclosing sides as $b = 6$ and $c = 5$ .

State the rule (SSS). Three sides and no angle means use the angle form:

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

Substitute the given values.

\cos A = \frac{6^2 + 5^2 - 9^2}{2 \times 6 \times 5} = \frac{36 + 25 - 81}{60} = \frac{-20}{60} \approx -0.3333

Invert to find the angle.

A = \cos^{-1}(-0.3333) \approx 109\degree

Check. The numerator $b^2 + c^2 - a^2$ is negative, which predicts an obtuse angle, and $109\degree$ is indeed obtuse. The largest angle of the garden bed is about $109\degree$ .

core3 marks

In triangle

ABC

, the side

a = 12

cm and the side

b = 15

cm meet at an included angle

C = 95^\circ

. Find the length of the third side

c

, correct to two decimal places.

Show worked solution →

Label the SAS triangle. The angle $C = 95^\circ$ sits between the two known sides $a = 12$ and $b = 15$ , so the side $c$ opposite $C$ comes from the side form.

Mind the obtuse angle. Because $95^\circ$ is obtuse, $\cos 95^\circ \approx -0.0872$ is negative, so the last term is added rather than subtracted:

c^2 = 12^2 + 15^2 - 2 \times 12 \times 15 \times \cos 95^\circ = 144 + 225 - 360 \times (-0.0872) = 400.38

Take the square root.

c \approx \sqrt{400.38} \approx 20.01 \text{ cm}

Check. The angle is just past a right angle, so $c$ should be a little longer than the Pythagoras value $\sqrt{12^2 + 15^2} \approx 19.21$ cm, and $20.01$ cm is, as expected. The third side is about $20.01$ cm.

core3 marks

Two straight fences meet at a corner post at an angle of

115\degree

. One fence runs

45

m and the other runs

70

m from the post. Find the distance between the two far ends of the fences, correct to two decimal places.

Show worked solution →

Label the SAS triangle. Put the corner post at vertex $C$ . The two fences are the sides $a = 45$ and $b = 70$ , and the angle between them is the included angle $C = 115\degree$ . The distance we want, $c$ , is opposite that angle.

State and substitute the side form. Because $115\degree$ is obtuse, $\cos 115\degree \approx -0.4226$ is negative, so the last term is added:

c^2 = 45^2 + 70^2 - 2 \times 45 \times 70 \times \cos 115\degree = 2025 + 4900 - 6300 \times (-0.4226) = 9587.50

Take the square root.

c \approx \sqrt{9587.50} \approx 97.92 \text{ m}

Check. With a wide obtuse angle the far ends should be well separated, close to the sum of the fences ( $115$ m) rather than their difference ( $25$ m), and $97.92$ m sits sensibly in that range. The ends are about $97.92$ m apart.

core3 marks

A triangle has sides of length

13

cm,

14

cm and

15

cm. Find the size of the smallest angle, correct to the nearest degree.

Show worked solution →

Identify the smallest angle. The smallest angle lies opposite the shortest side, which is the $13$ cm side. Put that side as $a = 13$ and the enclosing sides as $b = 14$ and $c = 15$ .

State the rule (SSS).

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

Substitute the given values.

\cos A = \frac{14^2 + 15^2 - 13^2}{2 \times 14 \times 15} = \frac{196 + 225 - 169}{420} = \frac{252}{420} = 0.6000

Invert to find the angle.

A = \cos^{-1}(0.6000) \approx 53\degree

Check. The numerator is positive, so the angle is acute, and $53\degree$ is the smallest of the three angles because it faces the shortest side. The smallest angle is about $53\degree$ .

core4 marks

A bushwalker leaves a campsite on a bearing of

070\degree

and walks

18

km, then changes course to a bearing of

150\degree

and walks a further

24

km. Find the straight-line distance from the campsite to the walker's finishing point, correct to two decimal places.

Show worked solution →

Find the interior angle at the bend: The bearing turns by $150\degree - 070\degree = 80\degree$ . The interior angle of the triangle at the turning point is the supplement, $180\degree - 80\degree = 100\degree$ , because the new leg makes an $80\degree$ angle with the straight-ahead direction.
Set up the SAS triangle: The two legs $18$ km and $24$ km enclose the $100\degree$ angle, and the unknown $c$ is the distance from the campsite.
State and substitute the side form: Since $\cos 100\degree \approx -0.1736$ , the last term is added:

c^2 = 18^2 + 24^2 - 2 \times 18 \times 24 \times \cos 100\degree = 324 + 576 - 864 \times (-0.1736) = 1050.03

Take the square root.

c \approx \sqrt{1050.03} \approx 32.40 \text{ km}

Check. The result lies between the difference of the legs ( $6$ km) and their sum ( $42$ km), as it must, so the walker finishes about $32.40$ km from the campsite.

exam5 marks

A triangular paddock

PQR

has two sides meeting at corner

P

: the side

PQ = 95

m and the side

PR = 120

m, with an included angle of

78\degree

P

. (a) Find the length of the remaining side

QR

, correct to two decimal places. (b) Hence find the area of the paddock, correct to the nearest square metre.

Show worked solution →

Part (a), label the SAS triangle. The included angle $P = 78\degree$ sits between the two known sides $PQ = 95$ and $PR = 120$ , so the side $QR$ opposite $P$ comes straight from the side form.

Substitute and solve for $QR$ . Using $\cos 78\degree \approx 0.2079$ :

QR^2 = 95^2 + 120^2 - 2 \times 95 \times 120 \times \cos 78\degree = 9025 + 14400 - 22800 \times 0.2079 = 18684.61

QR \approx \sqrt{18684.61} \approx 136.69 \text{ m}

Part (b), choose the area formula. We already know two sides and the angle between them, so the included-angle area formula is the quickest route:

\text{Area} = \tfrac{1}{2} \times PQ \times PR \times \sin P = \tfrac{1}{2} \times 95 \times 120 \times \sin 78\degree

Substitute. Using $\sin 78\degree \approx 0.9781$ :

\text{Area} \approx \tfrac{1}{2} \times 95 \times 120 \times 0.9781 \approx 5575 \text{ m}^2

Check. The third side $QR \approx 136.69$ m is longer than each given side, which fits an angle near $78\degree$ opening the triangle wide, and the area is a little under the right-angled maximum of $\tfrac{1}{2} \times 95 \times 120 = 5700$ m $^2$ , exactly as expected when the angle is just under $90\degree$ .

exam5 marks

A ship sails from port

A

to port

B

on a bearing of

035\degree

, a distance of

60

km. At

B

it changes course to a bearing of

110\degree

and sails

90

km to port

C

. Find the direct distance from

A

C

, correct to two decimal places.

Show worked solution →

Find the interior angle at $B$: The bearing changes by $110\degree - 035\degree = 75\degree$ , so the interior angle of triangle $ABC$ at the bend is the supplement, $180\degree - 75\degree = 105\degree$ .
Set up the SAS triangle: The two legs $AB = 60$ km and $BC = 90$ km enclose the $105\degree$ angle at $B$ , and the unknown is the side $AC$ opposite $B$ .
State and substitute the side form: Because $105\degree$ is obtuse, $\cos 105\degree \approx -0.2588$ is negative and the final term is added:

AC^2 = 60^2 + 90^2 - 2 \times 60 \times 90 \times \cos 105\degree = 3600 + 8100 - 10800 \times (-0.2588) = 14495.25

Take the square root.

AC \approx \sqrt{14495.25} \approx 120.40 \text{ km}

Check. The distance lies between the difference of the legs ( $30$ km) and their sum ( $150$ km), and the obtuse turn pushes it towards the upper end, so $AC \approx 120.40$ km is sensible.

exam4 marks

A triangle has sides of length

7

8

m and

12

m. (a) Show that the angle opposite the

12

m side is obtuse. (b) Find that angle, correct to the nearest minute.

Show worked solution →

Part (a), set up the angle form. Put the side opposite the wanted angle as $a = 12$ , and the two enclosing sides as $b = 7$ and $c = 8$ :

\cos A = \frac{b^2 + c^2 - a^2}{2 b c} = \frac{7^2 + 8^2 - 12^2}{2 \times 7 \times 8} = \frac{49 + 64 - 144}{112} = \frac{-31}{112} \approx -0.2768

Read the sign. The numerator $b^2 + c^2 - a^2 = -31$ is negative, so $\cos A$ is negative and the angle must lie between $90\degree$ and $180\degree$ . The angle opposite the $12$ m side is therefore obtuse.

Part (b), invert to find the angle.

A = \cos^{-1}(-0.2768) \approx 106.0685\degree

Convert the decimal part to minutes. The whole-degree part is $106\degree$ , and $0.0685\degree \times 60 \approx 4'$ , so

A \approx 106\degree 04'

Check. The angle is obtuse as shown in part (a), and $106\degree$ is the largest angle of the triangle because it faces the longest side, which is consistent. The angle is about $106\degree 04'$ .

exam6 marks

A surveyor standing at station

S

sights two boundary markers

M

and

T

. The distance

SM = 110

m, the distance

ST = 140

m, and the angle

MST

between the two lines of sight is

62^\circ

. (a) Find the direct distance

MT

between the two markers, correct to two decimal places. (b) Hence find the size of the angle

SMT

at marker

M

, correct to the nearest degree.

Show worked solution →

Part (a), label the SAS triangle. The angle $S = 62^\circ$ sits between the two known sides $SM = 110$ and $ST = 140$ , so the side $MT$ opposite $S$ comes from the side form.

Substitute and solve for $MT$ . Using $\cos 62^\circ \approx 0.4695$ :

MT^2 = 110^2 + 140^2 - 2 \times 110 \times 140 \times \cos 62^\circ = 12100 + 19600 - 30800 \times 0.4695 = 17240.28

MT \approx \sqrt{17240.28} \approx 131.30 \text{ m}

Part (b), set up the angle form at $M$ . Now all three sides are known, so use the angle form with the wanted angle's opposite side as $ST = 140$ , and the enclosing sides $SM = 110$ and $MT \approx 131.30$ :

\cos M = \frac{SM^2 + MT^2 - ST^2}{2 \times SM \times MT} = \frac{110^2 + 131.30^2 - 140^2}{2 \times 110 \times 131.30} \approx 0.3372

Invert to find the angle.

M = \cos^{-1}(0.3372) \approx 70^\circ

Check. The three angles should sum to $180^\circ$ . With $S = 62^\circ$ and $M \approx 70^\circ$ , the angle at $T$ would be about $48^\circ$ , and $62 + 70 + 48 = 180$ , so the working is consistent. The angle $SMT$ is about $70^\circ$ .

Practise this

Sources & how we know this

Mathematics Standard Stage 6 Syllabus (2017) — NESA
Mathematics Standard HSC exam papers — NESA