D1.6 Solve two-dimensional right and oblique triangle problems

Ontario Grade 11 Mathematics

Trigonometric Functions

Right Triangles, the Sine Law, and the Cosine Law in Action

Two-dimensional triangle problems appear everywhere: a surveyor measuring a plot of land, a navigator plotting a course, or a designer calculating the reach of a ramp. In Grade 10 you learned to solve right triangles using the primary trigonometric ratios (sine, cosine, tangent) and the Pythagorean theorem. In this lesson you will extend those tools to oblique triangles — triangles that contain no right angle — using the Sine Law and the Cosine Law. By the end, you will be able to look at any triangle problem, choose the correct tool, and work through it step by step.

What you will learn

Decide whether to use a right-triangle ratio, the Sine Law, or the Cosine Law based on the information given.
Apply the Sine Law to find unknown sides and angles in oblique triangles.
Apply the Cosine Law to find unknown sides and angles in oblique triangles.
Set up and solve multi-step triangle problems drawn from real-world two-dimensional situations.
Check that answers are reasonable using triangle properties such as the largest side opposing the largest angle.

Prerequisite Bridge: Right-Triangle Trigonometry

Before working with oblique triangles, make sure you are comfortable with the three primary ratios for a right triangle. In a right triangle with an acute angle

\theta

, the side opposite

\theta

is called the opposite side, the side next to

\theta

(that is not the hypotenuse) is the adjacent side, and the longest side (across from the right angle) is the hypotenuse.

The three ratios are

\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}

,

\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}

, and

\tan\theta = \frac{\text{opposite}}{\text{adjacent}}

. To isolate an unknown side you rearrange the ratio; to find an unknown angle you apply the inverse trig function, for example

\theta = \sin^{-1}\!\left(\frac{\text{opposite}}{\text{hypotenuse}}\right)

.

These ratios only work when you can identify a right angle. As soon as the triangle has no right angle, you need one of the two laws introduced in this lesson.

SOH-CAH-TOA applies only to right triangles.
Use inverse trig ( $\sin^{-1}$ , $\cos^{-1}$ , $\tan^{-1}$ ) to find unknown angles.
All angles in any triangle add to CAD 180°.
The Pythagorean theorem $a^2 + b^2 = c^2$ also requires a right angle.

The Sine Law

An oblique triangle has three sides — labelled

a

,

b

,

c

— and three angles — labelled

A

,

B

,

C

. By convention, side

a

is opposite angle

A

, side

b

is opposite angle

B

, and side

c

is opposite angle

C

. The Sine Law states that the ratio of each side to the sine of its opposite angle is the same for all three pairs in a triangle.

You use the Sine Law when you know: (1) two angles and any one side (AAS or ASA), or (2) two sides and an angle that is opposite one of those sides (SSA). In situation (2), be aware that SSA can sometimes produce two valid triangles — always check whether the computed angle, when subtracted from 180°, also produces a valid triangle that fits the given information.

To find a missing side, place the unknown side ratio on the left and solve by cross-multiplying. To find a missing angle, isolate the sine of that angle, then apply the inverse sine function. Remember that

\sin^{-1}

on a calculator always returns a value between CAD 0° and CAD 90°, so you must decide whether the obtuse supplement is also a valid solution.

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

Use the Sine Law when you know an angle and its opposite side, plus one more piece of information.
Cross-multiply to isolate an unknown side; use $\sin^{-1}$ to isolate an unknown angle.
SSA may give two solutions — always check both.
Angles in a triangle sum to CAD 180°, so you can often find the third angle first.

The Cosine Law

The Cosine Law connects all three sides of a triangle to one of its angles. It is the tool to reach for when you know: (1) three sides and want an angle (SSS), or (2) two sides and the angle between them and want the third side (SAS). In both cases, no angle–opposite-side pair is initially available, so the Sine Law cannot start the solution.

There are three equivalent forms of the Cosine Law — one for each angle. To find side

a

, use the form that has

\cos A

on the right. To find angle

A

from three known sides, rearrange that same form to isolate

\cos A

, then apply

\cos^{-1}

.

Because

\cos^{-1}

returns values from CAD 0° to CAD 180°, it correctly handles obtuse angles — a major advantage over starting with the Sine Law in SSS or SAS problems. Once you find one angle with the Cosine Law, you can find a second with the Sine Law and the third from the angle sum.

a^2 = b^2 + c^2 - 2bc\cos A

Use the Cosine Law for SAS (finding the third side) and SSS (finding any angle).
Each form of the Cosine Law involves one angle and all three sides.
Rearrange to isolate $\cos A$ when finding angles from three known sides.
After finding one element with the Cosine Law, switch to the Sine Law for efficiency.

Choosing the Right Tool

Before you write a single equation, study the triangle diagram and list what is given (sides or angles) and what is unknown. Your choice of method follows directly from that list. Right-triangle ratios work only when a right angle is confirmed. The Sine Law works when at least one angle–opposite-side pair is fully known. The Cosine Law works when that pair is not available.

A common strategy for multi-step problems is to split a complex figure into simpler triangles, solve each triangle in sequence, and carry forward any sides or angles found as you go. Carry at least four significant figures through intermediate steps and round only your final answer.

Right triangle: use SOH-CAH-TOA or the Pythagorean theorem.
Known angle–opposite-side pair plus one extra: use the Sine Law.
SAS or SSS with no known angle–opposite-side pair: use the Cosine Law.
Multi-step: divide the figure into triangles and solve in sequence.
Round only at the final step to avoid accumulated rounding error.

Reasonableness Checks

After solving, always verify that your answer makes geometric sense. The largest side of a triangle must be opposite the largest angle, and all three angles must add to exactly CAD 180°. If a computed side is longer than the sum of the other two sides, something has gone wrong — a triangle cannot be formed in that case.

For applied problems, re-read the question and confirm that the size of your answer matches the real-world context. A flagpole calculated to be 400 m tall or a boat bearing of 270° when the problem says it is heading roughly north-east should prompt you to recheck signs or calculator mode.

Largest side is always opposite the largest angle.
All three interior angles must sum to CAD 180°.
Check that every side satisfies the triangle inequality.
Confirm your answer fits the real-world context of the problem.

Choosing Your Triangle Tool

Given information	Unknown	Tool to use
Right angle confirmed, two sides or one side and one acute angle	Remaining side or angle	SOH-CAH-TOA / Pythagorean theorem
Two angles and any side (AAS or ASA)	A side	Sine Law
Two sides and the angle opposite one of them (SSA)	An angle or side	Sine Law (check for two solutions)
Two sides and the included angle (SAS)	The third side	Cosine Law
All three sides (SSS)	Any angle	Cosine Law

Worked example

Sine Law: Finding a Side Across a Pond

A surveyor needs to find the distance across a pond from point

P

to point

Q

. She sets up a third point

R

on dry land. She measures

PR = 74\text{ m}

, angle

P = 61°

, and angle

R = 48°

. Find the distance

PQ

to the nearest metre.

Find the missing angle
The three interior angles of any triangle sum to CAD 180°. Angles $P$ and $R$ are given, so angle $Q$ equals $180° - 61° - 48°$ .
$Q = 180° - 61° - 48° = 71°$
Identify the known angle–opposite-side pair
Side $PR$ (length $74\text{ m}$ ) is opposite angle $Q = 71°$ . That gives a complete ratio $\frac{PR}{\sin Q}$ to anchor the Sine Law. The unknown side $PQ$ is opposite angle $R = 48°$ .
Write the Sine Law for these two pairs
Set up the two relevant ratios equal to each other. Side $PQ$ is opposite $R = 48°$ , and side $PR = 74$ is opposite $Q = 71°$ .
$\frac{PQ}{\sin 48°} = \frac{74}{\sin 71°}$
Isolate and evaluate $PQ$
Multiply both sides by $\sin 48°$ to isolate $PQ$ . Using a calculator: $\sin 48° \approx 0.7431$ and $\sin 71° \approx 0.9455$ . PQ = $\frac{74 × \sin 48°}{\sin 71°}$ = $\frac{74 × 0.7431}{0.9455}$ \approx 58.15 $\text{ m}$
Round and state the answer
Rounding to the nearest metre gives the distance across the pond. PQ \approx 58 $\text{ m}$

Answer: The distance

PQ

across the pond is approximately

58\text{ m}

.

Check: Check: The largest angle is

Q = 71°

, so the longest side should be

PR = 74\text{ m}

. Indeed

74 > 58

, and angle

P = 61°

is between

R = 48°

and

Q = 71°

, so the opposite side

QR

should be between

58\text{ m}

and

74\text{ m}

. Using the Sine Law:

QR = \frac{74 \sin 61°}{\sin 71°} \approx \frac{74 \times 0.8746}{0.9455} \approx 68.4\text{ m}

. This is between

58

and

74

, confirming the solution is consistent.

Worked example

Cosine Law: Finding the Distance Between Two Ships

Two ships leave the same harbour at the same time. Ship A travels

52\text{ km}

on a bearing and Ship B travels

39\text{ km}

on a different bearing, with an angle of CAD 118° between their paths. How far apart are the two ships when they stop? Round to the nearest kilometre.

Recognise the triangle type
We know two sides ( $52\text{ km}$ and $39\text{ km}$ ) and the angle between them (CAD 118°). This is a SAS situation with no known angle–opposite-side pair, so the Cosine Law is the correct starting tool.
Label the triangle and write the Cosine Law
Let $d$ be the unknown distance between the ships. Set $b = 52$ , $c = 39$ , and $A = 118°$ (the included angle). The Cosine Law gives $d^2$ directly.
$d^2 = 52^2 + 39^2 - 2(52)(39)\cos 118°$
Evaluate the squared terms
Calculate each part separately to reduce errors. $52^2 = 2704$ and $39^2 = 1521$ , so $52^2 + 39^2 = 4225$ .
$52^2 + 39^2 = 4225$
Evaluate the cosine term
Using a calculator, $\cos 118° \approx -0.4695$ . Note the negative value — this is expected because CAD 118° is obtuse. Multiplying: $2 \times 52 \times 39 \times (-0.4695) = 4056 \times (-0.4695) \approx -1903.7$ .
$2(52)(39)\cos 118° \approx -1903.7$
Combine to find $d^2$
Substituting the computed values: $d^2 = 4225 - (-1903.7) = 4225 + 1903.7$ . When you subtract a negative number, you add.
$d^2 = 4225 + 1903.7 = 6128.7$
Take the square root and round
Take the positive square root (a distance cannot be negative) to find $d$ .
$d = \sqrt{6128.7} \approx 78.3\text{ km}$
State the final answer
Rounding to the nearest kilometre gives the distance between the two ships.
$d \approx 78\text{ km}$

Answer: The two ships are approximately

78\text{ km}

apart.

Check: Check: The included angle is obtuse (CAD 118°), so the side opposite it must be the longest side in the triangle. Our answer of

78\text{ km}

is longer than both

52\text{ km}

and

39\text{ km}

, which is geometrically correct. Also verify with the triangle inequality:

52 + 39 = 91 > 78

✓,

52 + 78 = 130 > 39

✓,

39 + 78 = 117 > 52

✓.

Common mistakes and how to avoid them

Using SOH-CAH-TOA on an oblique triangle that has no right angle.

Correction: Check for a right angle first. If none exists, use the Sine Law or the Cosine Law instead.

Forgetting that the SSA case (two sides and a non-included angle) can produce two different valid triangles.

Correction: After finding angle

B

with

\sin^{-1}

, also test

B' = 180° - B

and verify whether it produces a valid triangle whose angles sum to CAD 180°.

Subtracting instead of adding when

\cos A

is negative (obtuse angle) in the Cosine Law, treating the term

-2bc\cos A

as always negative.

Correction: Substitute the actual negative value of

\cos A

and then apply the subtraction sign in the formula carefully: subtracting a negative number gives addition.

Rounding intermediate values to two decimal places too early, causing the final answer to be off by several units.

Correction: Keep at least four significant figures in every intermediate calculation and round only the final stated answer.

Applying the Cosine Law with sides labelled incorrectly so that side

a

is not opposite angle

A

.

Correction: Always label vertices and their opposite sides consistently before writing any equation: side

a

is directly across from vertex

A

.

Lesson summary

Right-triangle trigonometry (SOH-CAH-TOA and the Pythagorean theorem) applies only when a right angle is present.
The Sine Law, $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ , is used when at least one angle–opposite-side pair is fully known (AAS, ASA, or SSA).
The Cosine Law, $a^2 = b^2 + c^2 - 2bc\cos A$ , is used for SAS (finding a side) and SSS (finding an angle).
In the SSA case, always check whether the supplement of the computed angle produces a second valid triangle.
Carry four or more significant figures through all intermediate steps and round only the final answer.
Verify every solution using the angle-sum property ( $A + B + C = 180°$ ), the largest-side-vs-largest-angle rule, and the triangle inequality.

Check your understanding

Question 1

A triangle has sides

a = 9\text{ cm}

,

b = 12\text{ cm}

, and the angle between them is

C = 54°

. Which tool should you use first to find side

c

?

SOH-CAH-TOA, because one angle is known.
The Sine Law, because two sides are known.
The Cosine Law, because two sides and the included angle are known (SAS).
The Pythagorean theorem, because two sides are known.

Show answer and explanation

The Cosine Law, because two sides and the included angle are known (SAS).

Knowing two sides and the angle between them is the SAS case. No angle–opposite-side pair is available, so the Sine Law cannot start the solution. The Cosine Law handles SAS directly:

c^2 = 9^2 + 12^2 - 2(9)(12)\cos 54°

.

Question 2

In triangle

XYZ

, angle

X = 35°

, angle

Y = 80°

, and side

y = 20\text{ cm}

(opposite angle

Y

). What is the value of side

x

to the nearest centimetre?

$x \approx 12\text{ cm}$
$x \approx 16\text{ cm}$
$x \approx 20\text{ cm}$
$x \approx 23\text{ cm}$

Show answer and explanation

x \approx 12\text{ cm}

Using the Sine Law:

\frac{x}{\sin 35°} = \frac{20}{\sin 80°}

, so

x = \frac{20 \times \sin 35°}{\sin 80°} = \frac{20 \times 0.5736}{0.9848} \approx \frac{11.47}{0.9848} \approx 11.6\text{ cm}

, which rounds to

12\text{ cm}

.

Question 3

You compute angle

B = 42°

using the Sine Law in an SSA problem where side

b = 15

and side

a = 18

. What must you check next?

Whether $B$ should be rounded to the nearest degree.
Whether the supplement $B' = 138°$ also produces a valid triangle.
Whether to switch to the Cosine Law to confirm the answer.
Whether the triangle is actually a right triangle.

Show answer and explanation

Whether the supplement

B' = 138°

also produces a valid triangle.

In an SSA problem,

\sin^{-1}

only returns values from CAD 0° to CAD 90°. The supplement

180° - 42° = 138°

has the same sine value. You must check whether using

B' = 138°

still allows all three angles to sum to CAD 180° and whether the resulting triangle is geometrically possible.

Question 4

A triangle has all three sides known:

p = 7

,

q = 10

,

r = 13

. Which expression correctly isolates

\cos P

so you can find angle

P

?

$\cos P = \frac{7^2 - 10^2 - 13^2}{2(10)(13)}$
$\cos P = \frac{10^2 + 13^2 - 7^2}{2(10)(13)}$
$\cos P = \frac{7^2 + 10^2 - 13^2}{2(7)(10)}$
$\cos P = \frac{p}{\sin P}$

Show answer and explanation

\cos P = \frac{10^2 + 13^2 - 7^2}{2(10)(13)}

The Cosine Law form for side

p

is

p^2 = q^2 + r^2 - 2qr\cos P

. Rearranging to isolate

\cos P

:

\cos P = \frac{q^2 + r^2 - p^2}{2qr} = \frac{10^2 + 13^2 - 7^2}{2(10)(13)} = \frac{100 + 169 - 49}{260} = \frac{220}{260} \approx 0.846

, giving

P \approx 32°

.

Key terms

Oblique triangle: A triangle that contains no right angle; all three angles are either acute or one is obtuse.
Sine Law: The rule that in any triangle, each side divided by the sine of its opposite angle gives the same value: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ .
Cosine Law: The rule that relates all three sides and one angle of any triangle, for example $a^2 = b^2 + c^2 - 2bc\cos A$ .
Included angle: The angle that sits directly between two known sides of a triangle.
SSA (ambiguous case): The situation where two sides and a non-included angle are known; it can produce zero, one, or two valid triangles.
Inverse trigonometric function: A function such as $\sin^{-1}$ , $\cos^{-1}$ , or $\tan^{-1}$ that returns the angle whose trigonometric value equals a given number.
Triangle inequality: The rule that the sum of any two sides of a triangle must be greater than the third side.
Bearing: A direction measured as an angle in degrees, typically clockwise from north, used in navigation problems.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Mathematics (MCR3U), expectation D1.6. It is a study resource, not an official curriculum publication.

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D1.6 · Solve two-dimensional right and oblique triangle problems

Ontario Grade 11 Mathematics

Right Triangles, the Sine Law, and the Cosine Law in Action

What you will learn

Prerequisite Bridge: Right-Triangle Trigonometry

The Sine Law

The Cosine Law

Choosing the Right Tool

Reasonableness Checks

Choosing Your Triangle Tool

Sine Law: Finding a Side Across a Pond

Cosine Law: Finding the Distance Between Two Ships

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Question 4

Key terms

Continue through MCR3U

About this lesson