VertexMath · MPM2D · Free worksheet

Acute Triangle Trigonometry Review

Try each problem first. You don’t need to finish everything.

Core path ~55 min Full sheet ~1 h 45 min

Recommended: K1–K5, then one A and one T problem

K = Knowledge & Understanding · A = Application · T = Thinking · Boss = Challenge

vertexmath.ca Hints, answers & more worksheets

Quick reference

Sine law: $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$ . Each side is paired with the angle opposite it. To find an angle, put the sines on top: $\dfrac{\sin B}{b}=\dfrac{\sin A}{a}$ .
Cosine law: $c^2=a^2+b^2-2ab\cos C$ , where $C$ is the angle between $a$ and $b$ . For an angle: $\cos C=\dfrac{a^2+b^2-c^2}{2ab}$ .
Which law? A side and its opposite angle are both known (AAS, ASA, SSA): sine law. SAS or SSS: cosine law.
Area $=\frac12ab\sin C$ . Bearings are measured clockwise from north and written with three digits ( $040^\circ$ ).
Most common trap: in the cosine law, multiply $2ab\cos C$ before subtracting. Use degree mode, keep full calculator values until the end, then round lengths to one decimal place and angles to the nearest tenth of a degree.

Show your work on separate paper. Show your work in the dot-grid space.

Core Skills (K)

Sine Law · Cosine Law · Choosing the Law

K1 Sine Law: Find the Side

Round lengths to one decimal place.

a)
In $\triangle ABC$ , $\angle A=52^\circ$ , $\angle B=71^\circ$ and $a=12$ cm. Find $b$ .
b)
In $\triangle PQR$ , $\angle P=48^\circ$ , $\angle Q=63^\circ$ and $PQ=15$ cm. Find $QR$ .
c)
Find $x$ .

Stuck? Peek at a hint

Pair each side with the angle opposite it. In (b), find $\angle R$ first: it is the angle opposite $PQ$ .

K2 Sine Law: Find the Angle

In acute $\triangle ABC$ , $a=11$ cm, $b=9$ cm and $\angle A=64^\circ$ .

a)
Find $\angle B$ .
b)
Find $\angle C$ .
c)
Find $c$ .

Stuck? Peek at a hint

Write the sine law with the sines on top, $\dfrac{\sin B}{b}=\dfrac{\sin A}{a}$ , then multiply both sides by $b$ .

K3 Cosine Law: Find the Side

a)
In $\triangle DEF$ , $d=8$ cm, $e=11$ cm and $\angle F=47^\circ$ . Find $f$ .
b)
Find $x$ .
c)
In $\triangle XYZ$ , $x=9$ , $y=12$ and $\angle Z=90^\circ$ . Use the cosine law to find $z$ .

Stuck? Peek at a hint

Work out $2ab\cos C$ as one number first, then subtract it from $a^2+b^2$ .

K4 Cosine Law: Find the Angle

In $\triangle ABC$ , $a=7$ cm, $b=8$ cm and $c=10$ cm. Round angles to the nearest tenth of a degree.

a)
Find $\angle C$ .
b)
Find $\angle A$ .
c)
Another triangle has sides $a=5$ , $b=6$ and $c=7$ . Determine $\cos C$ as a fraction in lowest terms.

Stuck? Peek at a hint

Use $\cos C=\dfrac{a^2+b^2-c^2}{2ab}$ : the side opposite the angle you want is the one subtracted.

K5 Which Law First?

The table gives the known measures of four acute triangles $ABC$ . Lengths are in centimetres.

$\text{Triangle 1}$	$\angle A=40^\circ$	$\angle B=75^\circ$	$a=10$
$\text{Triangle 2}$	$a=6$	$b=9$	$\angle C=58^\circ$
$\text{Triangle 3}$	$a=6$	$b=7$	$c=8$
$\text{Triangle 4}$	$a=12$	$b=10$	$\angle A=70^\circ$

a)
For each triangle, state the case (AAS, SAS, SSS or SSA) and the law you would use first.
b)
Triangle 2: find $c$ .
c)
Triangle 4: find $\angle B$ .

Stuck? Peek at a hint

Look for a side together with the angle opposite it: with such a pair the sine law can start, without one it cannot.

Real World (A)

Bearings · Area · Multi-Step Problems

A1 Back to Tobermory

A sailboat leaves Tobermory harbour ( $T$ ) and sails 14 km on a bearing of $040^\circ$ to a buoy $P$ . It then turns and sails 9 km on a bearing of $150^\circ$ to point $Q$ .

a)
Sketch the route and find $\angle TPQ$ , the angle of the triangle at $P$ .
b)
How far is the boat from the harbour, and on what bearing must it sail to return directly? Round the bearing to the nearest degree.
c)
Another boat's log reads: "14 km on one bearing, then 9 km on another; finished 25 km from the harbour." What can you conclude about the log?

Stuck? Peek at a hint

Draw a north line at $P$ parallel to the one at $T$ , and mark the $40^\circ$ angle it makes with $PT$ .

A2 Community Garden

A community garden in Mississauga is a triangle. Two of its sides run along paths and measure 18 m and 25 m, with an angle of $62^\circ$ between them.

a)
Find the area of the garden.
b)
The third side needs a fence that costs $14.50 per metre. Find the length of that side and the cost of the fence.
c)
The city wants a $250$ m² garden along the same two paths, with sides of 18 m and 25 m, changing only the angle between them. Is this possible? Explain.

Stuck? Peek at a hint

Use Area $=\frac12ab\sin C$ , where $C$ is the angle between the two known sides.

A3 Signal on the Highway

A straight highway starts at Junction $J$ and runs east. A cell tower $C$ stands 10 km from $J$ , and segment $JC$ makes a $60^\circ$ angle with the highway. The tower's signal reaches 12 km.

a)
Find the shortest distance from the tower to the highway.
b)
A driver heading east loses the signal $x$ km from $J$ . Use the cosine law to write an equation in $x$ , then solve it.
c)
Which solution answers the question, and what would the other one mean? State the stretch of highway, measured from $J$ , where a driver has signal.

Stuck? Peek at a hint

In the triangle formed by $J$ , $C$ and the point where the signal ends, the 12 km side is opposite the $60^\circ$ angle.

Think It Through (T)

Error Analysis · Counterexample · Build a Triangle

T1 Spot the Flaw

In $\triangle ABC$ , $a=7$ cm, $b=9$ cm and $\angle C=52^\circ$ . A student writes:
" $c^2=7^2+9^2-2(7)(9)\cos52^\circ=130-126\cos52^\circ=4\cos52^\circ\approx2.46$ , so $c\approx1.6$ cm."

a)
Without redoing the calculation, explain how you can tell that $1.6$ cm is impossible.
b)
Identify the error and find $c$ .
c)
A second student says: "The cosine law is Pythagoras with an extra term, so $c$ should be about $\sqrt{7^2+9^2}\approx11.4$ cm." For which angle $C$ is $c^2=a^2+b^2$ true? Explain why $c<11.4$ cm here.

Stuck? Peek at a hint

For (a), ask how short the third side of a triangle can be when the other two are 7 cm and 9 cm.

T2 Angles and Opposite Sides

a)
A student claims: "If one angle of a triangle is twice another, the side opposite it is twice as long." Give a counterexample.
b)
In $\triangle ABC$ , $a=6$ cm and $\angle A=40^\circ$ . Choose a length $b$ for which no triangle exists, and explain what goes wrong in the sine law.
c)
Build an acute triangle with whole-number sides, not all equal, that has a $60^\circ$ angle. Use the cosine law to show it works.

Stuck? Peek at a hint

In (a), the sine law pairs each side with the sine of its opposite angle. In (c), $\cos60^\circ=\frac12$ turns the cosine law into $c^2=a^2+b^2-ab$ .

Boss Level

Optional challenge

Bisect the Angle

In $\triangle ABC$ , $AB=8$ , $AC=5$ and $\angle A=60^\circ$ . The bisector of $\angle A$ meets $BC$ at $D$ .

a)
Find $BC$ and the exact area of $\triangle ABC$ .
b)
Find the exact length of $AD$ .
c)
Find $BD$ and $DC$ . Is $D$ the midpoint of $BC$ ?

Stuck? Peek at a hint

For (b), the areas of $\triangle ABD$ and $\triangle ACD$ add up to the area of $\triangle ABC$ , and each has a $30^\circ$ angle at $A$ .

Answer key

Final answers only. Try first, then check.

Core Skills

a) $b\approx14.4$ cm
b) $QR\approx11.9$ cm (using $\angle R=69^\circ$ )
c) $x\approx20.5$ m

a) $\angle B\approx47.3^\circ$
b) $\angle C\approx68.7^\circ$
c) $c\approx11.4$ cm

a) $f\approx8.1$ cm
b) $x\approx12.7$ cm
c) $z=15$ (the term $2xy\cos90^\circ$ is $0$ )

a) $\angle C\approx83.3^\circ$
b) $\angle A\approx44.0^\circ$
c) $\cos C=\dfrac{1}{5}$

a) 1: AAS, sine law. 2: SAS, cosine law. 3: SSS, cosine law. 4: SSA, sine law ( $a$ and $\angle A$ are a side with its opposite angle).
b) $c\approx7.7$ cm
c) $\angle B\approx51.5^\circ$

Real World

a) $\angle TPQ=70^\circ$
b) $TQ\approx13.8$ km; bearing $\approx258^\circ$ ( $Q$ lies on a bearing of about $078^\circ$ from $T$ )
c) It is wrong: after legs of 14 km and 9 km the boat is at most $14+9=23$ km from the harbour. The cosine law would need $\cos\theta=\frac{14^2+9^2-25^2}{2(14)(9)}\approx-1.38$ , and no angle has that cosine.

a) $\approx198.7$ m²
b) $\approx22.9$ m; about $332.70
c) No: the area is $\frac12(18)(25)\sin\theta=225\sin\theta$ , and $\sin\theta\le1$ , so the largest possible area is $225$ m² (at $90^\circ$ ). $\sin\theta=\frac{250}{225}\approx1.11$ has no solution.

a) $10\sin60^\circ\approx8.7$ km
b) $12^2=x^2+10^2-2(10)(x)\cos60^\circ$ , so $x^2-10x-44=0$ ; $x=5\pm\sqrt{69}$ , so $x\approx13.3$ or $x\approx-3.3$
c) $x\approx13.3$ km. The other solution, $-3.3$ , is a point 3.3 km west of $J$ , where there is no highway. Signal from $J$ to about 13.3 km: $0\le x\le13.3$ .

Think It Through

a) The sum of two sides is greater than the third: $c+7>9$ , so $c>2$ cm.
b) The student subtracted $130-126$ before multiplying by $\cos52^\circ$ . $c^2=130-126\cos52^\circ\approx52.43$ , so $c\approx7.2$ cm.
c) Only for $C=90^\circ$ , where $\cos C=0$ . For $C=52^\circ$ , $\cos C>0$ , so the positive amount $2ab\cos C$ is subtracted: $c^2<a^2+b^2$ , so $c<11.4$ cm.

a) e.g. angles $35^\circ$ , $70^\circ$ , $75^\circ$ : $\dfrac{\sin70^\circ}{\sin35^\circ}\approx1.64$ , not $2$ . Sides are proportional to the sines of the opposite angles, not to the angles.
b) Any $b>\dfrac{6}{\sin40^\circ}\approx9.3$ cm, e.g. $b=10$ cm: $\sin B=\dfrac{10\sin40^\circ}{6}\approx1.07$ , and no angle has a sine greater than $1$ .
c) e.g. $5$ , $7$ , $8$ : $5^2+8^2-2(5)(8)\cos60^\circ=89-40=49=7^2$ , so the angle opposite $7$ is $60^\circ$ (another: $8$ , $13$ , $15$ ).

Boss Level

Boss

a) $BC=7$ ; area $=10\sqrt3\approx17.3$
b) $AD=\dfrac{40\sqrt3}{13}\approx5.33$
c) $BD=\dfrac{56}{13}\approx4.31$ , $DC=\dfrac{35}{13}\approx2.69$ ; no ( $BD:DC=8:5=AB:AC$ )

Free cheat sheet

Acute Triangle Trigonometry Cheat Sheet

The whole unit on one page: every rule, a worked example for each idea, and the traps that cost marks.

Unit Summary

One printable page with every rule.

↓ Unit Summary PDF

Build your own

The same page with blanks: fill it in from memory, then check.

↓ Build-your-own PDF

Phone card

The essentials on one screen. Save it to your photos.

↓ Phone card (image)

Most testedIn acute $\triangle ABC$ , $b=9$ , $c=10$ and $\angle A=50^\circ$ . Find $a$ and $\angle B$ . (card 4)

1Sine law: a side

$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$

UseAAS: a side with its opposite angle. ASA: find the third angle first, so the known side has a pair.

Example $\angle A=50^\circ$ , $\angle B=65^\circ$ and $a=10$ give $b=10\sin65^\circ/\sin50^\circ\approx11.8$ .

TrapEach side pairs with the angle across from it, never an angle at one of its ends.

TrapSides are proportional to the sines: $a/A=b/B$ is false.

2Sine law: an angle

$\dfrac{\sin B}{b}=\dfrac{\sin A}{a}$

UsePut the sines on top, then use $\sin^{-1}$ . Keep full calculator values; round at the end.

Example $a=13$ , $b=10$ , $\angle A=72^\circ$ : $\sin B=10\sin72^\circ/13$ , so $\angle B\approx47.0^\circ$ .

Trap $\sin B=b\sin A/a$ , not $a\sin A/b$ : that can exceed $1$ and give a calculator error.

3Cosine law

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

$\cos C=\dfrac{a^2+b^2-c^2}{2ab}$

UseSAS: the third side. SSS: an angle. $c$ is the side opposite $C$ . At $C=90^\circ$ it becomes $c^2=a^2+b^2$ .

ExampleSides $6$ and $8$ , $50^\circ$ between: $c^2=100-96\cos50^\circ\approx38.3$ , so $c\approx6.2$ .

TrapSides $6$ , $8$ , $50^\circ$ between: $c^2=100-96\cos50^\circ$ , not $(100-96)\cos50^\circ$ . Multiply first.

4Using both laws

UseName the case first. SAS: cosine law for the third side, then the sine law for an angle. Angles add to $180^\circ$ .

Example $b=9$ , $c=10$ , $\angle A=50^\circ$ : $a^2=181-180\cos50^\circ$ , $a\approx8.1$ ; $\sin B=9\sin50^\circ/a$ , $\angle B\approx58.6^\circ$ .

TrapSAS has no side with its opposite angle, so the sine law can't start: use the cosine law first.

5Bearings

UseMeasured clockwise from north, with three digits ( $030^\circ$ ). Draw a north line at every point; the north lines are parallel.

Example $030^\circ$ , then $160^\circ$ : the angle inside the triangle at the turn is $180^\circ-(160^\circ-30^\circ)=50^\circ$ .

Trap $030^\circ$ then $160^\circ$ : the turn angle is $50^\circ$ , not the difference, $130^\circ$ .

6Area and chained triangles

$\text{Area}=\tfrac12ab\sin C$

Use $C$ is the angle between $a$ and $b$ . Two triangles: solve one, then carry the shared side into the next.

ExampleSides $12$ and $15$ , $40^\circ$ between: $\tfrac12(12)(15)\sin40^\circ\approx57.9$ .

TrapUse $\sin C$ : $\tfrac12ab\cos C$ and $\tfrac12ab$ are wrong unless $C=90^\circ$ (then $\tfrac12ab$ works).

Which law?

A side and its opposite angle Sine law

Two sides and the angle between Cosine law

Three sides Cosine law

↓ Worksheet PDF ↓ Answer key PDF

Google Classroom Email

More free MPM2D worksheets

Other courses

All MPM2D resources → · All free resources →

What is VertexMath?

VertexMath is daily math practice for Ontario high school students, with a tutor that asks questions instead of giving answers. These worksheets are free, whether or not you use it.

Visit vertexmath.ca →