What is sign slip with an obtuse angle in the cosine rule?

For an angle between 90 and 180, cos is negative, so the term -2abcos C becomes a positive number you add.

What is wrong calculator mode?

Maths Advanced triangle work is in degrees, not radians. A radian-mode answer is wildly wrong.

§-Syllabus dot point

NSWMaths AdvancedSyllabus dot point

How do the sine, cosine and area rules solve non-right-angled triangles, and how are they applied to elevation, depression, bearings and three-dimensional problems?

Solve problems using the sine rule, cosine rule and area rule, including angles of elevation and depression, bearings and three-dimensional applications

A focused answer to the HSC Maths Advanced dot point on triangle trigonometry. The sine rule and its ambiguous SSA case, the cosine rule in both side and angle forms, the area rule, angles of elevation and depression, compass and true bearings, and three-dimensional applications, with stage-by-stage diagrams and worked HSC-style examples.

Generated by Claude Opus 4.818 min answerUpdated 2026-06-21

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

For non-right-angled triangles, choose by the given information: a side paired with its opposite angle means the sine rule $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ (check the ambiguous case in SSA), two sides and the included angle (or three sides) means the cosine rule $c^2 = a^2 + b^2 - 2ab\cos C$ (rearranged to $\cos A = \frac{b^2+c^2-a^2}{2bc}$ for an angle), and the area is $\frac{1}{2}ab\sin C$ with the included angle. Angles of elevation and depression are equal alternate angles; bearings are clockwise from north in three figures, with the triangle's interior angle being the supplement at a bend or the difference at a station; and three-dimensional problems reduce to a flat right-angled triangle picked out of the solid. Keep the calculator in degrees and carry full precision to the last line.

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What this dot point is asking
The answer
How exam questions ask about triangle trigonometry

What this dot point is asking

NESA wants you to solve any triangle, not just right-angled ones, using the three rules on the reference sheet (the sine rule, the cosine rule, and the area rule $A = \frac{1}{2}ab\sin C$ ), and then apply them to the real geometry that the HSC asks about year after year: angles of elevation and depression, compass and true bearings, and three-dimensional figures such as prisms, towers and mountain peaks.

Note

Think about walking around a lake to work out how wide it is. You cannot swim across with a tape measure, but you can stand on the shore, pace out a couple of distances you can reach, and measure the angle you turn through - and that is enough to pin down the far side. Triangle trigonometry is exactly this trick: give it any two sides and an angle, or any three sides, and it hands back every missing length and angle of the triangle. The sine and cosine rules are the two tools that do the work, and once you can solve a triangle you can solve a surveyor's problem, a "how tall is that tower" problem, or a "what direction is the ship heading" problem, because each of those is just a triangle drawn on the ground or in the air. The whole job is spotting which triangle to draw and which rule fits the numbers you were given.

The rules themselves are printed on the reference sheet, so almost none of the marks are for recall. They are for two decisions and one habit. The first decision is which rule fits the information you have: a side paired with its opposite angle points to the sine rule, two sides with the angle wedged between them point to the cosine rule. The second decision, in three-dimensional and bearings work, is which triangle to pull out of the picture and which interior angle the words actually describe. The habit is keeping your calculator in degrees and carrying full precision to the last line. Get the rule, the triangle and the angle right and the arithmetic is a single calculator entry.

The answer

Every rule on this page refers to the same labelling: a triangle $ABC$ with side $a$ opposite angle $A$ , side $b$ opposite angle $B$ and side $c$ opposite angle $C$ . Fix this convention before anything else, because mispairing a side with the wrong angle is the most common way these questions are lost.

The sine rule

For any triangle $ABC$ ,

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}.

Flipped over, with the sines on top, it is more convenient when the unknown is an angle:

\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}.

Both forms say the same thing. The sine rule works whenever you can form a complete pair, a side and the angle directly opposite it, plus one more matching side or angle. That covers two configurations:

AAS / ASA (two angles and any side). Find the third angle from $A + B + C = 180\degree$ , then the side-on-top form gives any remaining side. Two angles fix the shape uniquely, so there is no ambiguity.
SSA (two sides and a non-included angle). The sine-on-top form gives the angle opposite the second known side. This is the configuration where the ambiguous case can appear.

A consequence worth carrying as a sanity check: the largest side always faces the largest angle.

The ambiguous case (SSA)

When you know two sides and an angle not between them, the data can fit two different triangles. The reason is purely about $\sin^{-1}$ : for any value $k$ with $0 < k < 1$ , two angles between $0\degree$ and $180\degree$ have that sine, an acute one $\sin^{-1}k$ and an obtuse one $180\degree - \sin^{-1}k$ . Geometrically, fix the known angle and one known side from its vertex, then swing the side opposite the known angle like a compass arc; depending on its length the arc can cut the base line at two points, one point, or none.

The safe routine for any SSA problem is: compute $\sin(\text{unknown}) = k$ , write down both $\sin^{-1}k$ and $180\degree - \sin^{-1}k$ , then test each against the angle-sum constraint $A + B + C = 180\degree$ and keep every one that survives. There is only one triangle when:

The known angle is obtuse or right. A triangle has at most one non-acute angle, so the obtuse alternative would force two, which is impossible.
The side opposite the known angle is at least as long as the other given side. The swung arc then reaches the base only once.
The obtuse alternative pushes the angle sum to $180\degree$ or beyond. Test it; if the known angle plus the obtuse candidate is $\geq 180\degree$ , discard the obtuse one.

The cosine rule

When you cannot form a complete pair, the cosine rule takes over. For a side from two sides and the included angle (SAS):

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

where $C$ is the angle between the two given sides $a$ and $b$ , and $c$ is the side opposite it. The same pattern rotates around the triangle ( $a^2 = b^2 + c^2 - 2bc\cos A$ , and so on). Rearranged to make an angle the subject, for three sides (SSS):

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

with $A$ opposite the side $a$ . The sign of the numerator tells you the type of angle at a glance: if $b^2 + c^2 - a^2 < 0$ the angle is obtuse, if it is zero the angle is exactly $90\degree$ , and if positive it is acute. When the included angle is a right angle, $\cos 90\degree = 0$ , the rule collapses to $c^2 = a^2 + b^2$ , so the cosine rule is just Pythagoras generalised to any triangle.

The area rule

The area of any triangle from two sides and the angle between them:

\text{Area} = \frac{1}{2}ab\sin C = \frac{1}{2}bc\sin A = \frac{1}{2}ac\sin B.

The single load-bearing word is included: the angle must be the one squeezed between the two sides you multiply. The formula is base times height in disguise, since dropping a perpendicular from one vertex gives height $a\sin C$ , and $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}b(a\sin C)$ .

Key fact

The three rules for any triangle $ABC$ , side $a$ opposite angle $A$ and so on:

Sine rule: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$ . Use it when you have a complete side-angle pair (AAS or SSA). Put the unknown on top; in SSA, check the ambiguous case.
Cosine rule: $c^2 = a^2 + b^2 - 2ab\cos C$ for a side (SAS), rearranged to $\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}$ for an angle (SSS).
Area rule: $\text{Area} = \dfrac{1}{2}ab\sin C$ , with $C$ the included angle.

The deciding question is always whether a side is paired with its opposite angle: if yes, sine rule; if the only angle you know sits between two known sides (or you know all three sides), cosine rule.

Angles of elevation and depression

An angle of elevation is measured upward from the horizontal to a line of sight; an angle of depression is measured downward from the horizontal. The two are linked by a fact that saves a step in almost every question: the angle of depression from a high point to a low point equals the angle of elevation from the low point back to the high point, because the two horizontal lines are parallel and the line of sight is a transversal, making them alternate angles.

In a single right-angled triangle, the elevation or depression angle $\theta$ , the horizontal distance $x$ and the vertical height $H$ are tied together by $\tan\theta = \frac{H}{x}$ . When the foot of the vertical is not directly accessible, or two observation points sit on different bearings, you build a second triangle on the ground and reach for the sine or cosine rule, which is exactly what the three-dimensional questions below do.

Compass and true bearings

A true bearing is the angle measured clockwise from north, written with three digits from $000\degree$ to $360\degree$ . The leading zeros are not decoration; they mark the number as a bearing and stop $075$ being misread as $750$ .

A compass bearing names a primary direction (N or S), an acute angle, then E or W. To convert, walk clockwise from north on the diagram: N $30\degree$ E is $030\degree$ ; S $40\degree$ W is past south by $40\degree$ , so $220\degree$ ; N $50\degree$ W is $360\degree - 50\degree = 310\degree$ . The single most useful rule for bearings problems concerns the interior angle of the triangle you build:

At a bend in a path (a ship changes course), the interior angle is the supplement of the change of direction, $180\degree - (\text{turn})$ .
At a station with two radii out to surveyed points, the interior angle is the difference of the two bearings.
A back-bearing reverses by $180\degree$ : the bearing of $A$ from $B$ is the bearing of $B$ from $A$ plus or minus $180\degree$ , because the two north arrows are parallel.

Every point in a bearings diagram needs its own north arrow, since a bearing is always read from the local north where you are standing.

Three-dimensional applications

Three-dimensional problems look intimidating but are solved by the same two rules, applied to a flat triangle you isolate from the solid. The whole skill is choosing the right plane:

Inside a rectangular prism, a vertical edge is perpendicular to the base, so a triangle joining a top vertex, a base vertex below it and a third base point is right angled at the base vertex. The horizontal leg is usually a diagonal across a base face, found by Pythagoras.
For a vertical pole or tower seen from two ground points, each elevation gives a right-angled triangle that fixes a horizontal distance ( $\frac{\text{height}}{\tan\theta}$ ), and the two ground points plus the foot form a triangle on the ground that you solve with the cosine rule, then read off a bearing.
For a mountain peak fixed from two bearings, the same plan applies: two vertical right triangles supply the two ground distances in terms of the height $h$ , and the horizontal triangle (with the included angle from the bearings) ties them together so you can solve for $h$ and then any bearing.

The recipe is always the same: name the right-angled triangles that contain the vertical, write each horizontal distance in terms of the height, then close the figure with the horizontal triangle and the cosine or sine rule.

How exam questions ask about triangle trigonometry

The wording tells you which rule and which triangle to reach for:

"Find the length of ... / find $x$ " with two angles and a side. AAS for the sine rule: find the third angle if needed, then side-on-top.
"Find the size of angle ..." with two sides and an angle opposite one of them. SSA for the sine rule, sine-on-top; then check the ambiguous case.
"... including the ambiguous case", "find ALL possible triangles", "is more than one triangle possible?" A direct ambiguous-case prompt: give both the acute and obtuse solutions that survive the angle-sum test.
"Find the length of ... / how far apart ..." with two sides and the angle between them. SAS for the cosine rule.
"Find the largest / smallest angle" with three sides. SSS for the cosine rule; target the angle opposite the longest or shortest side directly.
"Find the area" with two sides and the included angle. The area rule. With three sides only, find an angle by the cosine rule first; with two angles and a side, find a second side by the sine rule first.
"The angle of elevation / depression is ..." Draw the right-angled triangle, mark $\tan\theta = \frac{\text{height}}{\text{horizontal}}$ , and use the alternate-angle link if the angle is quoted at the other end.
"... on a bearing of ...", "find the bearing of ... from ...", "how far is ..." Convert bearings into the interior angle (supplement at a bend, difference at a station, $\pm 180\degree$ for a back-bearing), then it is an ordinary sine or cosine rule triangle.
"In the rectangular prism / pyramid shown ..." A three-dimensional problem: isolate the flat triangle containing the segment asked for, use the vertical edge as a right angle, and find any horizontal leg by Pythagoras.
"Show that ... = [given value]" A scaffold step: the value is handed to you so the next part can proceed, so set out the working that lands exactly on it.

Exam technique

Pick the rule from the given information before you compute: a side paired with its opposite angle means the sine rule, an angle wedged between two known sides (or three sides) means the cosine rule. State the rule, then substitute the actual numbers; markers award method marks for the correct setup even if the arithmetic slips. Put the unknown on top so the last step is a single multiply or divide, and for a cosine-rule side do not forget the final square root. Keep the calculator in degrees (DEG, not RAD) and carry full precision to the last line, rounding only the final answer to the precision asked. In every SSA question write both $\sin^{-1}k$ and $180\degree - \sin^{-1}k$ and test each; in bearings work give a north arrow at each point and three digits in the answer; in three-dimensional work redraw the single flat triangle you are using, off to the side, with its right angle marked. Finish with a sanity check: the longest side faces the largest angle.

Worked example

Find a side in an AAS triangle

In triangle $ABC$ , $A = 43\degree$ , $B = 61\degree$ and the side $a = 12.4$ cm (opposite $A$ ). Find the side $b$ .

Find the third angle (optional here). $C = 180\degree - 43\degree - 61\degree = 76\degree$ . You do not need $C$ for $b$ , but finding it is free and lets you check the longest-side rule at the end.

Set up the matched pairs. The side $a = 12.4$ is opposite $A = 43\degree$ , and the wanted side $b$ is opposite $B = 61\degree$ . Two complete pairs, so the sine rule links them with the unknown on top:

\frac{b}{\sin B} = \frac{a}{\sin A} \implies b = \frac{a\sin B}{\sin A} = \frac{12.4\sin 61\degree}{\sin 43\degree}.

Evaluate: Carrying full precision, $b \approx 15.90$ cm.
Sanity check: $B = 61\degree > A = 43\degree$ , so $b$ should exceed $a$ , and $15.90 > 12.4$ . Good.
Answer: $b \approx 15.90$ cm (to 2 decimal places).

Resolve an ambiguous SSA triangle

In triangle $ABC$ , $A = 29\degree$ , $a = 15$ cm and $b = 27$ cm. Find angle $B$ , then complete each possible triangle.

Apply the sine rule for the angle. The side $b = 27$ is opposite the unknown $B$ ; the side $a = 15$ is opposite the known $A = 29\degree$ . Sine on top:

\sin B = \frac{b\sin A}{a} = \frac{27\sin 29\degree}{15} \approx 0.8727.

Write down both candidate angles. Since $0 < 0.8727 < 1$ , two angles share this sine:

B_1 = \sin^{-1}(0.8727) \approx 60.8\degree, \qquad B_2 = 180\degree - 60.8\degree \approx 119.2\degree.

Test each against the angle sum: For $B_1$ : $A + B_1 = 29\degree + 60.8\degree = 89.8\degree < 180\degree$ , leaving $C_1 \approx 90.2\degree$ . Valid. For $B_2$ : $A + B_2 = 29\degree + 119.2\degree = 148.2\degree < 180\degree$ , leaving $C_2 \approx 31.8\degree$ . Also valid. Both survive, which is the ambiguous case in action ( $b > a$ with $A$ acute).
Complete each triangle with the sine rule: Using $\frac{c}{\sin C} = \frac{a}{\sin A}$ : for the first triangle, $c_1 = \frac{15\sin 90.2\degree}{\sin 29\degree} \approx 30.94$ cm; for the second, $c_2 = \frac{15\sin 31.8\degree}{\sin 29\degree} \approx 16.29$ cm.
Answer: $B \approx 60.8\degree$ (with $C \approx 90.2\degree$ , $c \approx 30.94$ cm) or $B \approx 119.2\degree$ (with $C \approx 31.8\degree$ , $c \approx 16.29$ cm). Both are geometrically valid.

Find a distance with the cosine rule after a change of course

A bushwalker leaves a hut and walks $6.2$ km, then turns and walks a further $4.5$ km. The change of direction makes the interior angle at the turn $115\degree$ . How far is the walker from the hut?

Confirm the configuration. The two legs $6.2$ and $4.5$ enclose the $115\degree$ angle, and the wanted distance is the side opposite it. This is SAS, so the cosine rule applies directly.

Substitute into the side form. With $\cos 115\degree \approx -0.4226$ (note the negative, so the last term is added):

c^2 = 6.2^2 + 4.5^2 - 2(6.2)(4.5)\cos 115\degree \approx 38.44 + 20.25 + 23.58 = 82.27.

Take the square root: $c \approx \sqrt{82.27} \approx 9.07$ km.
Sanity check: The straight-line distance should be less than the two legs added end to end ( $10.7$ km) but more than their difference ( $1.7$ km), and $9.07$ sits between.
Answer: the walker is about $9.07$ km from the hut (to 2 decimal places).

Find the largest angle with the cosine rule (SSS)

A triangular garden bed has sides $8$ m, $11$ m and $15$ m. Find its largest angle.

Target the right angle. The largest angle faces the longest side, $15$ m, so put $a = 15$ and the two enclosing sides as $b = 8$ , $c = 11$ .

Use the angle form of the cosine rule.

\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{8^2 + 11^2 - 15^2}{2(8)(11)} = \frac{64 + 121 - 225}{176} = \frac{-40}{176} \approx -0.2273.

Invert. Since the numerator is negative, $\cos A < 0$ and the angle is obtuse: $A = \cos^{-1}(-0.2273) \approx 103.1\degree$ .

Answer: the largest angle is about $103.1\degree$ (to 1 decimal place).

Find an area, then convert to hectares

A triangular paddock has two sides of $35$ m and $42$ m meeting at an angle of $68\degree$ . Find its area in square metres, then in hectares.

Apply the area rule. The $68\degree$ is the included angle between the two sides, so:

\text{Area} = \frac{1}{2}ab\sin C = \frac{1}{2}(35)(42)\sin 68\degree \approx 681.5 \text{ m}^2.

Convert to hectares. Since $1$ ha $= 10\,000$ m $^2$ , divide: $681.5 \div 10\,000 \approx 0.0681$ ha.

Answer: about $681.5$ m $^2$ , or about $0.0681$ ha (to 4 decimal places).

Find a horizontal distance from an angle of depression

A lighthouse keeper stands at a point $O$ that is $80$ m above sea level and sees a boat $B$ at an angle of depression of $12\degree$ . The boat then sails directly towards the lighthouse until the angle of depression is $20\degree$ . How far did the boat travel?

Use the alternate-angle link: The angle of depression from $O$ equals the angle of elevation from the boat, so each right-angled triangle has the boat's horizontal distance $x$ as the base, the height $80$ as the opposite side, and $\tan\theta = \frac{80}{x}$ , i.e. $x = \frac{80}{\tan\theta}$ .
First position ( $12\degree$ ): $x_1 = \frac{80}{\tan 12\degree} \approx 376.4$ m.
Second position ( $20\degree$ ): $x_2 = \frac{80}{\tan 20\degree} \approx 219.8$ m.
Subtract: The boat moved closer by $x_1 - x_2 \approx 376.4 - 219.8 = 156.6$ m.
Answer: the boat travelled about $156.6$ m (to 1 decimal place).

Find a distance and a bearing on a two-leg course

A yacht leaves harbour $P$ and sails $18$ km on a bearing of $052\degree$ to $A$ , then $25$ km on a bearing of $145\degree$ to $B$ . Find the distance $PB$ and the bearing of $B$ from $P$ .

Find the interior angle at $A$ . The course turns from $052\degree$ to $145\degree$ , a change of $93\degree$ . The interior angle of the triangle at the bend is the supplement, $180\degree - 93\degree = 87\degree$ .

Cosine rule for $PB$ . The two legs $18$ and $25$ enclose the $87\degree$ angle, and $PB$ is opposite it:

PB^2 = 18^2 + 25^2 - 2(18)(25)\cos 87\degree \approx 901.90, \qquad PB \approx \sqrt{901.90} \approx 30.03 \text{ km}.

Sine rule for the angle at $P$ . The side opposite $\angle APB$ is the second leg $AB = 25$ :

\sin(\angle APB) = \frac{25\sin 87\degree}{30.03} \approx 0.8313, \qquad \angle APB \approx 56.23\degree.

Convert to a bearing. Adding this to the first bearing, the bearing of $B$ from $P$ is $052\degree + 56.23\degree \approx 108.2\degree$ .

Answer: $PB \approx 30.03$ km and $B$ is on a bearing of about $108\degree$ from $P$ .

Find an angle inside a rectangular prism (2023 HSC Q22)

In a rectangular prism, $AD = 7$ cm, $AE = 8$ cm and $EF = 6$ cm, where $AE$ is a vertical edge. The point $M$ is the midpoint of $CD$ . Find $\angle AEM$ , to the nearest degree.

Isolate the right-angled triangle: Since $AE$ is vertical it is perpendicular to the base, so triangle $AEM$ is right angled at $A$ , with the vertical leg $AE = 8$ and the horizontal leg $AM$ lying in the base.
Find the horizontal leg $AM$: The base is a rectangle with $AB = DC = EF = 6$ and $AD = 7$ . $M$ is the midpoint of $CD$ , so $DM = \frac{1}{2}(6) = 3$ . The segment $AM$ runs across the base from $A$ to $M$ , with horizontal offsets $AD = 7$ and $DM = 3$ , so by Pythagoras $AM = \sqrt{7^2 + 3^2} = \sqrt{58} \approx 7.616$ cm.
Find the angle: In right triangle $AEM$ , the angle at $E$ has opposite side $AM$ and adjacent side $AE$ :

\tan(\angle AEM) = \frac{AM}{AE} = \frac{\sqrt{58}}{8} \approx 0.9520, \qquad \angle AEM \approx 43.59\degree.

Answer: $\angle AEM \approx 44\degree$ (to the nearest degree), matching the NESA solution.

Find a peak height and bearing from two angles of elevation (2025 HSC Q29)

A peak $T$ stands directly above $O$ . $Y$ is due east of $O$ with the angle of elevation of $T$ from $Y$ equal to $60\degree$ . $F$ is $4$ km south-west of $Y$ with the angle of elevation of $T$ from $F$ equal to $45\degree$ ; $O$ , $Y$ and $F$ are on level ground. Let the height be $h$ . Show that $OY = \frac{h}{\sqrt{3}}$ , find $h$ , and find the bearing of $O$ from $F$ .

Set up the first right triangle: $TO$ is vertical, so triangle $TOY$ is right angled at $O$ with $\tan 60\degree = \frac{h}{OY}$ . Therefore $OY = \frac{h}{\tan 60\degree} = \frac{h}{\sqrt{3}}$ , as required.
Set up the second right triangle: Triangle $TOF$ is right angled at $O$ with $\tan 45\degree = \frac{h}{OF}$ , so $OF = h$ (since $\tan 45\degree = 1$ ).
Find the included angle on the ground: Looking from $Y$ , the direction to $O$ is due west (because $Y$ is due east of $O$ ), and the direction to $F$ is south-west, which is $45\degree$ west of south. The angle between due west and south-west at $Y$ is $135\degree$ , so $\angle OYF = 135\degree$ .
Apply the cosine rule in triangle $OYF$: The sides are $OF = h$ (opposite the angle at $Y$ ), $OY = \frac{h}{\sqrt{3}}$ and $YF = 4$ :

h^2 = \left(\frac{h}{\sqrt{3}}\right)^2 + 4^2 - 2\cdot\frac{h}{\sqrt{3}}\cdot 4\cdot\cos 135\degree.

Simplifying gives the exact solution $h = \sqrt{30} - \sqrt{6} \approx 3.03$ km (to 2 decimal places).

Find the bearing of $O$ from $F$ . Place $O$ at the origin. $Y$ is due east at $OY = \frac{h}{\sqrt{3}} \approx 1.748$ km, and $F$ is a further $4$ km south-west of $Y$ , which is $\frac{4}{\sqrt{2}} \approx 2.828$ km west and $2.828$ km south of $Y$ . So $F$ sits about $1.748 - 2.828 = -1.080$ km east (i.e. $1.080$ km west) and $2.828$ km south of $O$ . Seen from $F$ , the point $O$ is therefore to the north and east. The bearing is $\tan^{-1}\!\left(\frac{1.080}{2.828}\right) \approx 20.9\degree$ , i.e. about $021\degree$ .

Answer: $OY = \frac{h}{\sqrt{3}}$ ; $h \approx 3.03$ km; the bearing of $O$ from $F$ is about $021\degree$ , all matching the NESA solution.

Common traps

Reaching for the wrong rule: A side paired with its opposite angle calls for the sine rule; an angle wedged between two known sides, or three sides with no angle, calls for the cosine rule. Decide from the given information before substituting.
Mispairing sides and angles: Side $a$ faces angle $A$ . Always sketch and label; this is the single most common sine-rule and cosine-rule error.
Dropping the second SSA solution: $\sin B = k$ has two solutions on $0\degree$ to $180\degree$ . Write both and test each against the angle sum before discarding either.
Sign slip with an obtuse angle in the cosine rule: For an angle between $90\degree$ and $180\degree$ , $\cos$ is negative, so the term $-2ab\cos C$ becomes a positive number you add.
Forgetting the square root, or the units: After the cosine rule gives $c^2$ , take the positive square root for $c$ , and write areas in squared units.
Wrong calculator mode: Maths Advanced triangle work is in degrees, not radians. A radian-mode answer is wildly wrong.
Mis-reading a bearing as the interior angle: The interior angle of the triangle is the supplement of a course change at a bend, or the difference of two bearings at a station, not the raw bearing. Convert before using a rule, and reverse a back-bearing by $180\degree$ .
Solving the wrong triangle in 3D: Redraw the single flat triangle you need, off to the side, and mark the right angle the vertical edge creates before computing.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2023 HSC Q223 marksIn the rectangular prism shown,

AD = 7

cm,

AE = 8

cm and

EF = 6

cm. The point

M

is the midpoint of

CD

. Find

\angle AEM

, to the nearest degree.

Show worked answer →

The vertical edge $AE = 8$ is perpendicular to the base $ABCD$ , so triangle $AEM$ is right angled at $A$ . The other leg $AM$ lies flat in the base.

$M$ is the midpoint of $CD$ , so $DM = \frac{1}{2}DC = \frac{1}{2}(6) = 3$ . Then $AM$ is the diagonal of the base rectangle from $A$ to $M$ , with horizontal sides $DM = 3$ and $AD = 7$ :

$AM = \sqrt{3^2 + 7^2} = \sqrt{58} \approx 7.616$ cm.

In right triangle $AEM$ , $\tan(\angle AEM) = \frac{AM}{AE} = \frac{\sqrt{58}}{8}$ , so $\angle AEM = \tan^{-1}\!\left(\frac{\sqrt{58}}{8}\right) \approx 43.59\degree$ , which rounds to $44\degree$ .

Markers reward identifying the right angle at $A$ , finding $AM$ as a base diagonal with $DM = 3$ , and the final angle to the nearest degree.

2024 HSC Q204 marksA vertical tower

TC

40

m high. The point

A

is due east of the base

C

; the angle of elevation to the top

T

from

A

35\degree

. A second point

B

is on a different bearing from the tower; the angle of elevation to

T

from

B

30\degree

. The points

A

and

B

are

100

m apart. (a) Show that

AC = 57.13

m, correct to 2 decimal places. (b) Find the bearing of

B

from

C

, to the nearest degree.

Show worked answer →

(a) In right triangle $TCA$ , $\tan 35\degree = \frac{TC}{AC} = \frac{40}{AC}$ , so $AC = \frac{40}{\tan 35\degree} \approx 57.13$ m.

(b) Similarly $BC = \frac{40}{\tan 30\degree} \approx 69.28$ m. Now use triangle $ABC$ on the ground, where $AB = 100$ , $AC \approx 57.13$ and $BC \approx 69.28$ . By the cosine rule for the angle at $C$ :

$\cos(\angle ACB) = \frac{AC^2 + BC^2 - AB^2}{2 \cdot AC \cdot BC} \approx -0.2447$ , so $\angle ACB \approx 104.16\degree$ .

$A$ is due east of $C$ , so $A$ is on bearing $090\degree$ from $C$ , and $B$ lies clockwise (south) of that direction, giving the bearing of $B$ from $C$ as $090\degree + 104.16\degree \approx 194\degree$ .

Markers reward the right-triangle step for each distance, the cosine rule for $\angle ACB$ , and combining it with the $090\degree$ direction of $A$ to get the bearing.

2025 HSC Q297 marks

T

is the peak of a mountain and

O

is directly below it.

Y

is due east of

O

with angle of elevation of

T

from

Y

equal to

60\degree

F

4

km south-west of

Y

with angle of elevation of

T

from

F

equal to

45\degree

;

O

Y

F

are on level ground. (a) Let the height be

h

. Show that

OY = \frac{h}{\sqrt{3}}

. (b) Find

h

, correct to 2 decimal places. (c) Find the bearing of

O

from

F

, to the nearest degree.

Show worked answer →

(a) In right triangle $TOY$ (right angle at $O$ ), $\tan 60\degree = \frac{h}{OY}$ , so $OY = \frac{h}{\tan 60\degree} = \frac{h}{\sqrt{3}}$ .

(b) In right triangle $TOF$ , $\tan 45\degree = \frac{h}{OF}$ , so $OF = h$ . The angle $OYF$ at $Y$ is $135\degree$ (the east direction $YO$ runs west from $Y$ , and $F$ is south-west of $Y$ , $45\degree$ off south, giving $135\degree$ between $YO$ and $YF$ ). By the cosine rule in triangle $OYF$ with $OF = h$ , $OY = \frac{h}{\sqrt{3}}$ , $YF = 4$ :

$h^2 = \left(\frac{h}{\sqrt{3}}\right)^2 + 4^2 - 2 \cdot \frac{h}{\sqrt{3}} \cdot 4 \cdot \cos 135\degree$ . Solving gives $h = \sqrt{30} - \sqrt{6} \approx 3.03$ km.

(c) With $O$ at the origin, $Y$ due east at distance $\frac{h}{\sqrt{3}} \approx 1.748$ km and $F$ a further $4$ km south-west of $Y$ , $F$ sits about $1.080$ km west and $2.828$ km south of $O$ . The bearing of $O$ from $F$ is therefore in the north-east quadrant: $\tan^{-1}\!\left(\frac{1.080}{2.828}\right) \approx 21\degree$ , i.e. about $021\degree$ .

Markers reward each right-triangle relation, the $135\degree$ included angle, the cosine rule solved for $h$ , and the bearing from the displacement of $O$ relative to $F$ .

Practice questions

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

foundation2 marksIn triangle

ABC

A = 38\degree

B = 95\degree

and side

a = 20

cm (opposite

A

). Find the length of side

b

, correct to 2 decimal places.

Show worked solution →

Choose the rule. The side $a = 20$ is opposite the known $A = 38\degree$ , and $b$ is opposite the known $B = 95\degree$ , so two complete side-angle pairs are available: this is the sine rule.

Put the unknown on top and substitute.

\frac{b}{\sin B} = \frac{a}{\sin A} \quad\Rightarrow\quad b = \frac{a\sin B}{\sin A} = \frac{20\sin 95\degree}{\sin 38\degree}.

Evaluate, carrying full precision.

b = \frac{20 \times 0.99619}{0.61566} \approx 32.36 \text{ cm}.

Marker's note: one mark for the correct sine-rule setup with the unknown on top, one for $b \approx 32.36$ cm. A sanity check earns confidence, not marks: $B > A$ , so $b$ should exceed $a = 20$ , and it does.

foundation2 marksA triangular sail has two edges of length

13

m and

17

m meeting at an angle of

49\degree

. Find its area, correct to 1 decimal place.

Show worked solution →

Choose the rule. Two sides and the included angle between them point straight to the area rule.

Substitute into $\text{Area} = \frac{1}{2}ab\sin C$ . Here the included angle is $C = 49\degree$ :

\text{Area} = \frac{1}{2}(13)(17)\sin 49\degree = 110.5\sin 49\degree.

Evaluate.

\text{Area} = 110.5 \times 0.75471 \approx 83.4 \text{ m}^2.

Marker's note: one mark for the area-rule substitution with the included angle, one for $83.4$ m $^2$ with squared units. Using an angle that is not wedged between the two sides is the usual slip.

foundation3 marksA triangular garden bed has sides of

6

9

m and

11

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

Show worked solution →

Choose the rule and target angle. Three sides and no angle means the cosine rule; the largest angle faces the longest side, so set $a = 11$ with the enclosing sides $b = 6$ and $c = 9$ .

Use the angle form of the cosine rule.

\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{6^2 + 9^2 - 11^2}{2(6)(9)} = \frac{36 + 81 - 121}{108} = \frac{-4}{108} \approx -0.03704.

Invert. The negative cosine flags an obtuse angle:

A = \cos^{-1}(-0.03704) \approx 92.12\degree \approx 92\degree.

Marker's note: one mark for choosing the cosine rule and targeting the angle opposite the longest side, one for the correct substitution, one for $A \approx 92\degree$ . A positive-numerator slip that hides the obtuse angle loses the final mark.

core3 marksA ship leaves port

P

and sails

40

km, then changes course and sails a further

65

km to reach

Q

. The interior angle of the triangle at the turning point is

72\degree

. Find the direct distance

PQ

, correct to 2 decimal places.

Show worked solution →

Choose the rule. The two legs $40$ and $65$ enclose the $72\degree$ angle at the bend, and $PQ$ is the side opposite it: two sides and the included angle, so the cosine rule for a side.

Substitute into the side form.

PQ^2 = 40^2 + 65^2 - 2(40)(65)\cos 72\degree = 1600 + 4225 - 5200\cos 72\degree.

Evaluate the cosine term, then take the square root. With $\cos 72\degree \approx 0.30902$ ,

PQ^2 \approx 5825 - 1606.89 = 4218.11, \qquad PQ \approx \sqrt{4218.11} \approx 64.95 \text{ km}.

Marker's note: one mark for the cosine-rule setup, one for the value $PQ^2 \approx 4218.11$ , one for $PQ \approx 64.95$ km after the square root. Forgetting the final square root (leaving $4218.11$ ) is the classic error.

core4 marksIn triangle

ABC

A = 35\degree

a = 9

cm and

b = 12

cm. Find all possible values of angle

B

, and for each state the corresponding angle

C

, correct to 1 decimal place.

Show worked solution →

Choose the rule. A side paired with its opposite angle ( $a$ with $A$ ) plus a second side $b$ is the SSA configuration, so the sine rule with sine on top, and the ambiguous case must be checked.

Solve for $\sin B$ . The side $b = 12$ is opposite the unknown $B$ :

\sin B = \frac{b\sin A}{a} = \frac{12\sin 35\degree}{9} \approx 0.76477.

Write down both candidate angles. Since $0 < 0.76477 < 1$ , two angles share this sine:

B_1 = \sin^{-1}(0.76477) \approx 49.9\degree, \qquad B_2 = 180\degree - 49.9\degree \approx 130.1\degree.

Test each against the angle sum. For $B_1$ : $A + B_1 = 35\degree + 49.9\degree = 84.9\degree < 180\degree$ , so $C_1 \approx 95.1\degree$ . Valid. For $B_2$ : $A + B_2 = 35\degree + 130.1\degree = 165.1\degree < 180\degree$ , so $C_2 \approx 14.9\degree$ . Also valid. Both survive, because $b > a$ with $A$ acute.

Marker's note: one mark for the sine-rule setup, one for $B_1 \approx 49.9\degree$ , one for identifying the second solution $B_2 \approx 130.1\degree$ , one for the two matching values of $C$ after the angle-sum test. Reporting only the acute solution throws away the second triangle and a mark.

core4 marksA field surveyor walks the boundary of a triangular reserve and records the three straight edges in a table as

120

150

m and

190

m. (a) Find the largest interior angle of the reserve, correct to the nearest degree. (b) Hence find the area of the reserve, correct to the nearest square metre.

Show worked solution →

Part (a): read the data and pick the rule. The table gives all three sides and no angle, so the cosine rule applies. The largest angle faces the longest edge, so set $a = 190$ with the enclosing sides $b = 120$ and $c = 150$ :

\cos A = \frac{120^2 + 150^2 - 190^2}{2(120)(150)} = \frac{14400 + 22500 - 36100}{36000} = \frac{800}{36000} \approx 0.02222.

A = \cos^{-1}(0.02222) \approx 88.73\degree \approx 89\degree.

Part (b): use the area rule with the angle just found. The angle $A \approx 88.73\degree$ is the included angle between the two sides $b = 120$ and $c = 150$ :

\text{Area} = \frac{1}{2}bc\sin A = \frac{1}{2}(120)(150)\sin 88.73\degree \approx 9000 \times 0.99975 \approx 8998 \text{ m}^2.

Marker's note: one mark for the cosine-rule setup on the longest side, one for $A \approx 89\degree$ , one for the area-rule substitution with the included angle, one for the area $\approx 8998$ m $^2$ . Carrying the unrounded angle $88.73\degree$ into part (b) avoids rounding drift in the area.

exam5 marksA yacht leaves harbour

H

and sails

18

km on a bearing of

032\degree

to a buoy

A

, then

27

km on a bearing of

146\degree

to a buoy

B

. (a) Show that the interior angle of triangle

HAB

A

66\degree

. (b) Find the distance

HB

, correct to 2 decimal places. (c) Find the bearing of

B

from

H

, correct to the nearest degree.

Show worked solution →

Part (a): turn the bearings into the interior angle at $A$ . The course changes from $032\degree$ to $146\degree$ , a turn of $146\degree - 32\degree = 114\degree$ . At the bend the interior angle of the triangle is the supplement of the turn:

\angle HAB = 180\degree - 114\degree = 66\degree,

as required.

Part (b): cosine rule for $HB$ . The two legs $HA = 18$ and $AB = 27$ enclose the $66\degree$ angle, and $HB$ is opposite it:

HB^2 = 18^2 + 27^2 - 2(18)(27)\cos 66\degree = 324 + 729 - 972\cos 66\degree.

HB^2 \approx 1053 - 395.47 = 657.53, \qquad HB \approx \sqrt{657.53} \approx 25.64 \text{ km}.

Part (c): sine rule for the angle at $H$ , then convert to a bearing. The side opposite $\angle AHB$ is $AB = 27$ :

\sin(\angle AHB) = \frac{27\sin 66\degree}{25.64} \approx 0.96204, \qquad \angle AHB \approx 74.20\degree.

The first leg $HA$ leaves $H$ on bearing $032\degree$ , and $B$ lies clockwise of that direction, so the bearing of $B$ from $H$ is

032\degree + 74.20\degree \approx 106\degree.

Marker's note: one mark for the supplement giving $\angle HAB = 66\degree$ in (a); one for the cosine-rule setup and one for $HB \approx 25.64$ km in (b); one for the sine-rule angle at $H$ and one for the bearing $106\degree$ in (c). Adding $\angle AHB$ to the wrong reference direction, or forgetting the three-figure form, loses the final mark.

exam5 marksA vertical radio tower has its base at

O

on level ground. From a point

P

due south of

O

the angle of elevation of the top of the tower is

28\degree

. From a point

Q

the angle of elevation is

22\degree

, and

Q

lies on a bearing of

070\degree

from

O

. The points

P

and

Q

are

180

m apart. Let the height of the tower be

h

metres. (a) Show that

OP = \frac{h}{\tan 28\degree}

and write down a similar expression for

OQ

. (b) Explain why the angle

\angle POQ = 110\degree

, and hence find the height

h

, correct to 2 decimal places.

Show worked solution →

Part (a): each elevation gives a right-angled triangle on the ground. Since the tower $OT$ is vertical, triangle $TOP$ is right angled at $O$ , so

\tan 28\degree = \frac{h}{OP} \quad\Rightarrow\quad OP = \frac{h}{\tan 28\degree}.

The same argument in triangle $TOQ$ gives

OQ = \frac{h}{\tan 22\degree}.

Part (b): find the ground angle from the bearings. Seen from $O$ , the point $P$ is due south, on bearing $180\degree$ , and $Q$ is on bearing $070\degree$ . The angle between these two directions at $O$ is

\angle POQ = 180\degree - 70\degree = 110\degree.

Apply the cosine rule in the ground triangle $POQ$ . The side opposite the $110\degree$ angle is $PQ = 180$ , with $OP = \frac{h}{\tan 28\degree}$ and $OQ = \frac{h}{\tan 22\degree}$ :

180^2 = \left(\frac{h}{\tan 28\degree}\right)^2 + \left(\frac{h}{\tan 22\degree}\right)^2 - 2\cdot\frac{h}{\tan 28\degree}\cdot\frac{h}{\tan 22\degree}\cos 110\degree.

Factor out $h^2$ and evaluate the bracket with $\tan 28\degree \approx 0.53171$ and $\tan 22\degree \approx 0.40403$ :

32400 = h^2\left(\frac{1}{0.53171^2} + \frac{1}{0.40403^2} - \frac{2\cos 110\degree}{0.53171 \times 0.40403}\right) \approx 12.8474\,h^2.

h^2 = \frac{32400}{12.8474} \approx 2522.30, \qquad h \approx \sqrt{2522.30} \approx 50.22 \text{ m}.

Marker's note: one mark for each right-triangle relation in (a); one for justifying $\angle POQ = 110\degree$ from the two bearings; one for the cosine-rule equation in $h^2$ ; one for solving to $h \approx 50.22$ m. Treating $\angle POQ$ as the raw bearing $070\degree$ , or dropping the negative sign of $\cos 110\degree$ , are the two ways this question is lost.