Explore the development of the sine law for acute triangles

Ontario Grade 10 Mathematics

Trigonometry

Develop the relationship between a triangle’s sides and their opposite angles

In a right triangle, sine compares a side with an angle. An acute triangle has all three angles smaller than a right angle, but it is not a right triangle. We can draw an altitude inside it to create right triangles. Comparing the sine ratios in those smaller triangles leads to the sine law.

What you will learn

Recall how sine is defined in a right triangle.
Use an altitude to form right triangles inside an acute triangle.
Explain how using the same altitude leads to the sine law.
Use the sine law to find a missing side in an acute triangle.

1. Grade 9 bridge: sine in a right triangle

In a right triangle, the hypotenuse is the side opposite the right angle. For another chosen angle, the opposite side is across from that angle. Sine is the ratio of the opposite side to the hypotenuse.

An altitude is a perpendicular line segment drawn from a vertex to the opposite side. It creates a right angle. In an acute triangle, the altitude from a vertex meets the opposite side inside the triangle, so it splits the triangle into two right triangles.

\sin(\text{angle})=\frac{\text{opposite side}}{\text{hypotenuse}}

Sine compares the opposite side with the hypotenuse in a right triangle.
An altitude creates right triangles where we can use sine.

2. Develop the relationship with an altitude

Name the vertices of an acute triangle

A

,

B

, and

C

. A vertex is a corner of the triangle. Use lowercase letters for the opposite sides: side

a

is opposite angle

A

, side

b

is opposite angle

B

, and side

c

is opposite angle C.

Draw an altitude from

C

to side

AB

, and call its length

h

. The altitude creates two right triangles. In the one beside angle

A

, the hypotenuse is

b

and the side opposite

A

is

h

. In the one beside angle

B

, the hypotenuse is

a

and the side opposite

B

is also h.

Using sine in each smaller triangle gives two expressions for the same height. Equating them connects sides

a

and

b

with their opposite angles. Drawing an altitude from

A

to side

BC

similarly connects sides

b

and c. Together, these relationships show that each side divided by the sine of its opposite angle has the same value.

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

The same altitude is the opposite side in both smaller right triangles.
A side must be paired with the sine of its opposite angle.
For an acute triangle, these altitudes lie inside the triangle.

3. Read and use the sine law

The equal ratios are called the sine law. Each ratio has a side on top and the sine of its opposite angle below. The law says these side-to-sine ratios are equal. It does not say that the sides or angles are all equal.

To find a missing side, set up equal ratios using a known side and its opposite angle and the missing side and its opposite angle. Rearrange the equation to isolate the unknown side. Use a calculator set to degrees when angles are given in degrees.

Before calculating, match each side with its opposite angle. Keep enough digits during the calculation and round only the final answer. A quick check is to compare the relative sizes of the angles and their opposite sides.

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

Match $a$ with $A$ , $b$ with $B$ , and $c$ with $C$ .
Use degree mode for angles written in degrees.
Check that the answer fits the side-angle information.

4. Independent practice

For each question, first write down the known opposite side-angle pair and the pair containing the unknown side. Then set up equal ratios. Keep calculator values unrounded until the end.

1. An acute triangle has

A=42^\circ

,

B=68^\circ

, and

a=9.0

cm. Find

b

to the nearest tenth of a centimetre.

2. An acute triangle has

A=35^\circ

,

B=80^\circ

, and

b=12

cm. Find

a

to the nearest tenth of a centimetre.

Use a known side together with its opposite angle.
Keep each side paired with its opposite angle in the equation.

The altitude creates two right-triangle sine relationships

Right triangle	Angle used	Opposite side	Hypotenuse	Sine relationship
Triangle beside angle A	$A$	$h$	$b$	$\sin A=\frac{h}{b}$
Triangle beside angle B	$B$	$h$	$a$	$\sin B=\frac{h}{a}$

Worked example

Find a side using opposite side-angle pairs

An acute triangle has

A=38^\circ

,

B=72^\circ

, and

a=7.5

cm. Find

b

to the nearest tenth of a centimetre.

Match opposite pairs
Side $a$ is opposite angle $A$ , and side $b$ is opposite angle $B$ . The known pair is $a$ and $A$ ; the missing side belongs with $B$ .
Set up equal ratios
Use the sine law with the two opposite pairs. This keeps each side matched with its opposite angle.
$\frac{a}{\sin A}=\frac{b}{\sin B}$
Isolate the unknown side
Substitute the known values. Multiply by $\sin 72^\circ$ to leave $b$ by itself.
$b=\frac{7.5\sin 72^\circ}{\sin 38^\circ}$
Calculate and round
Evaluate in degree mode. Round the final length to the nearest tenth of a centimetre.
$b\approx 11.6\text{ cm}$

Answer: Side

b

is approximately

11.6

cm.

Check: Angle

B

is larger than angle

A

, so its opposite side should be longer than side

a

. The result, about

11.6

cm, is greater than

7.5

cm.

Common mistakes and how to avoid them

Pairing side

a

with angle

B

because both letters name parts of the same triangle.

Correction: Pair each lowercase side with the matching uppercase opposite angle:

a

with

A

,

b

with

B

, and

c

with

C

.

Using the adjacent side instead of the opposite side in the sine ratio.

Correction: Identify the chosen angle first. Sine uses the side across from that angle divided by the hypotenuse.

Using a calculator in radian mode for angles given in degrees.

Correction: Set the calculator to degree mode before evaluating a sine with a degree symbol.

Rounding intermediate values too early.

Correction: Keep calculator values unrounded during the calculation. Round only the final answer as requested.

Lesson summary

In a right triangle, sine is the opposite side divided by the hypotenuse.
An altitude inside an acute triangle creates right triangles where sine can be used.
The same altitude gives two expressions for one height. Equating them helps develop the sine law.
In the sine law, each side is paired with the sine of its opposite angle.

Check your understanding

Question 1

In an acute triangle, which angle is opposite side

c

?

Angle $A$
Angle $B$
Angle $C$
The right angle

Show answer and explanation

Angle

C

The lowercase side letter matches the uppercase letter of its opposite angle. Side

c

is opposite angle

C

.

Question 2

A triangle has

A=40^\circ

,

B=65^\circ

, and

a=8

cm. Which equation can be used to find

b

?

$\frac{8}{\sin 40^\circ}=\frac{b}{\sin 65^\circ}$
$\frac{8}{\sin 65^\circ}=\frac{b}{\sin 40^\circ}$
$\frac{8}{\sin 40^\circ}=\frac{b}{\sin 40^\circ}$
$\frac{8}{\sin 65^\circ}=\frac{b}{\sin 65^\circ}$

Show answer and explanation

\frac{8}{\sin 40^\circ}=\frac{b}{\sin 65^\circ}

Side

a=8

is opposite

A=40^\circ

, and side

b

is opposite

B=65^\circ

. The first equation pairs both sides with their opposite angles.

Question 3

Why can an altitude help develop the sine law for an acute triangle?

It changes all three sides to the same length.
It creates right triangles where the sine ratio can be used.
It makes every angle a right angle.
It removes the need to match sides with opposite angles.

Show answer and explanation

It creates right triangles where the sine ratio can be used.

The altitude creates smaller right triangles. Their sine ratios use the same height, which can be equated to build the side-angle relationship.

Key terms

Acute triangle: A triangle whose three angles are each less than a right angle.
Altitude: A perpendicular line segment from a vertex to the opposite side.
Opposite side: The side across from a chosen angle.
Hypotenuse: The side opposite the right angle in a right triangle.
Sine law: The rule that each side divided by the sine of its opposite angle gives the same ratio in a triangle.

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Published by DoAssignment. This reviewed lesson follows Ontario Grade 10 Mathematics (MPM2D), study topic T7. It is a study resource, not an official curriculum publication.

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