Solve metric and imperial applications with sine or cosine law

Ontario Grade 11 Mathematics

Geometry and Trigonometry

Ontario Grade 11 MBF3C | Study-guide label: C2.4

Some distances are hard to measure directly, such as the distance across a field or between two points separated by an obstacle. A triangle can help you find them. In this lesson, you will use the sine law and cosine law in practical problems with metric and imperial measurements. You will review how triangle sides and angles match, learn when each law applies, and practise with two applications. When angles are given in degrees, set your calculator to degree mode.

What you will learn

Identify known sides and angles in a practical triangle problem.
Choose the sine law or cosine law based on the given information.
Solve metric and imperial applications and state answers with suitable units.

1. Review: triangle labels and measurements

A triangle has three sides and three interior angles. The angles add to

180^\circ

. A side is opposite the angle across from it. In a labelled triangle, side

a

is opposite angle

A

, side

b

is opposite angle

B

, and side

c

is opposite angle

C

.

A labelled sketch helps you keep track of which measurements go together. Draw the triangle, write the known side lengths and angles beside it, and mark the unknown. The sketch does not need to be to scale, but the labels must match the problem.

Metric lengths may be given in metres or kilometres. Imperial lengths may be given in feet or miles. Keep the side lengths in one unit within a calculation. The answer for an unknown side will use that same unit. Use degree mode when the angles are given in degrees.

A+B+C=180^\circ

Pair each side with the angle directly across from it.
Keep length units consistent within a calculation.
A sketch is useful even when it is not drawn to scale.

2. Choose the law that fits the information

The sine law connects each side with the sine of its opposite angle. A complete opposite pair means a side and the angle across from it are both known. When a problem gives such a pair, the sine law can help find another side or angle, as long as the other needed measurement is available.

The cosine law connects three sides with one of the triangle’s angles. Use it to find a side when you know two sides and the angle between them. The angle between two sides that meet is called the included angle. The cosine law can also be used when all three sides are known and you need an angle.

Before choosing, list the information in the problem and mark it on a sketch. A known opposite side-angle pair points to the sine law. Two known sides and their included angle point to the cosine law. Choose based on what is given, not on the order in which information appears.

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

A known opposite side-angle pair supports using the sine law.
Two sides and the angle between them support using the cosine law to find a side.
Match every side and angle by their labels before substituting.

3. Calculate, interpret, and check

For the cosine law, use the version that matches the unknown side. If the known sides are

a

and

b

, and their included angle is

C

, the unknown side opposite

C

is

c

. Substitute the measurements carefully. When the calculation gives

c^2

, take the positive square root: a length cannot be negative.

For the sine law, match each side with its opposite angle. Rearrange the relationship to isolate the unknown side or angle. If finding an angle, use the inverse sine function on your calculator and check that the result fits the triangle and the information given.

Keep extra calculator digits during intermediate calculations, then round as the question requests. State what the answer measures and include its unit. Check whether the result is reasonable compared with the other sides and the practical situation. If all three angles are known or found, they should total

180^\circ

.

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

Use the positive square root when a calculation gives the square of a side.
Keep full calculator values until the final rounding step.
Include a unit and check whether the result makes sense.

Worked example

Find a metric distance with the cosine law

A surveyor marks two points on one side of a field. The points are 6 km and 9 km from a third point. The angle between the two measured directions is

52^\circ

. Find the distance between the two marked points, to the nearest hundredth of a kilometre.

Identify the given sides and angle
The two known sides meet at the known angle, so $52^\circ$ is the included angle. The cosine law fits this information. Let $c$ be the distance opposite that angle.
$a=6,\quad b=9,\quad C=52^\circ$
Substitute into the cosine law
Use the cosine law with the two known sides and their included angle. Since the side lengths are in kilometres, the resulting length will also be in kilometres.
$c^2=6^2+9^2-2(6)(9)\cos 52^\circ$
Calculate and round
Evaluate the right side in degree mode, then take the positive square root because $c$ is a length. The unrounded result is about $7.1086$ km, which rounds to the nearest hundredth.
$c\approx 7.11\text{ km}$

Answer: The distance between the two marked points is about 7.11 km.

Check: The answer is longer than 6 km and shorter than 9 km. That is reasonable for the triangle formed by the two given sides and their included angle.

Worked example

Find an imperial distance with the sine law

Two observation points and a cabin form a triangle. From the cabin, the angles toward the observation points are

41^\circ

and

67^\circ

. The side opposite the

41^\circ

angle is 120 ft. Find the side opposite the

67^\circ

angle, to the nearest foot.

Identify the opposite pair
The given angle and the side opposite it form a complete opposite pair: $41^\circ$ and 120 ft. This information supports using the sine law to find the side opposite $67^\circ$ .
$A=41^\circ,\quad a=120\text{ ft},\quad B=67^\circ$
Set up the sine law
Pair each side with its opposite angle. Rearrange the sine law to make the unknown side $b$ the subject.
$b=\frac{120\sin 67^\circ}{\sin 41^\circ}$
Calculate and round
Evaluate the expression in degree mode. The unrounded result is about 168.37 ft. The measurements are in feet, so round the final length to the nearest foot in feet.
$b\approx 168\text{ ft}$

Answer: The side opposite the

67^\circ

angle is about 168 ft.

Check: The

67^\circ

angle is larger than the

41^\circ

angle, and the side opposite it is longer: about 168 ft compared with 120 ft.

Common mistakes and how to avoid them

Using the sine law without a known opposite side-angle pair.

Correction: Check for a known side and its opposite angle. If there is no such pair, check whether the cosine law fits the given information.

Matching a side with an angle beside it rather than the angle across from it.

Correction: Label a sketch and match each side with the angle directly opposite it.

Using the cosine law with the wrong angle.

Correction: For finding a side, use the angle between the two known sides.

Rounding too early or leaving off the unit.

Correction: Keep extra calculator digits until the final step, then round as requested and state the unit.

Lesson summary

Use the sine law when a known side and its opposite angle are available.
Use the cosine law for two sides and their included angle when finding the third side.
Label carefully, keep units consistent, and check that the final answer fits the triangle and situation.

Check your understanding

Question 1

A triangle has two known sides and the angle between them. Which law is the best choice to find the third side?

Sine law
Cosine law
The angle-sum rule only
Either law without checking the given information

Show answer and explanation

Cosine law

The cosine law uses two side lengths and their included angle to find the third side.

Question 2

In a triangle, angle

A=35^\circ

, angle

B=80^\circ

, and side

a=10

m is opposite angle

A

. Which expression finds side

b

?

$b=\frac{10\sin 80^\circ}{\sin 35^\circ}$
$b=\frac{10\sin 35^\circ}{\sin 80^\circ}$
$b=10+80-35$
$b=10\cos 80^\circ$

Show answer and explanation

b=\frac{10\sin 80^\circ}{\sin 35^\circ}

The sine law pairs

a=10

with its opposite angle

35^\circ

, and pairs

b

with

80^\circ

. Rearranging gives the first expression.

Question 3

A calculation uses side lengths measured in feet. What unit should the calculated side have?

Metres
Degrees
Feet
Square feet

Show answer and explanation

Feet

The sine and cosine laws give a length in the same unit as the side lengths used.

Key terms

Opposite side: The side across from a particular angle in a triangle.
Included angle: The angle formed by two sides that meet.
Sine law: A rule that relates each side of a triangle to the sine of its opposite angle.
Cosine law: A rule that relates the sides of a triangle to one of its angles.

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About this lesson and its review

Published by DoAssignment. This reviewed lesson follows Ontario Grade 11 Mathematics (MBF3C), expectation C2.4. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

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