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VCE 11 General 2023

11.02 Trigonometry ratios

Lesson

Trigonometric ratios

A ratio is a way of stating a mathematical relationship comparing two quantities and is often represented as a fraction. In a right-angled triangle the ratios of the sides are called trigonometric ratios.

The three basic trigonometric ratios are Sine, Cosine and Tangent. These names are often shortened to become sin, cos and tan respectively. They are given by the ratio of sides relative to an angle, the reference angle.

Trigonometric ratios can be used to find an unknown side-length or an unknown interior angle in a right-angled triangle.

The hypotenuse is the longest length in the triangle, which is always opposite the 90\degree angle.

The opposite side-length is opposite to the reference angle \theta.

The adjacent side-length is adjacent to the reference angle \theta.

This image shows the relationship of the sine, cosine and tangent ratios. Ask your teacher for more information.

\sin \theta =\dfrac{\text{Opposite }}{\text{Hypotenuse }}=\dfrac{\text{O}}{\text{H}} \qquad \cos \theta =\dfrac{\text{Adjacent }}{\text{Hypotenuse }}=\dfrac{\text{A}}{\text{H}} \qquad\tan \theta =\dfrac{\text{Opposite }}{\text{Adjacent }}=\dfrac{\text{O}}{\text{A}}

The mnemonic of SOH CAH TOA is helpful to remember the sides that apply to the different ratios of sine, cosine, and tangent.

Examples

Example 1

Consider the triangle in the figure. If \sin\theta=\dfrac{4}{5}:

Right triangle A B C with right angle A. Side A C has a length of x, A B has a length of 4, and B C has a length of 5.
a

Which angle is represented by \theta?

Worked Solution
Create a strategy

Use the sine ratio.

Apply the idea

Since 4 is in the numerator, the side length with length 4 is the opposite side and the hypotenuse has a length of 5. We need to look at the diagram and find the angle that is across from the opposite side with length of 4. So the angle is \angle BCA.

Reflect and check

We could have also named this angle \angle ACB, or even just \angle C since there is no ambiguity as to what that represents.

b

Find the value of \cos\theta.

Worked Solution
Create a strategy

To find the value of x, use the Pythagorean theorem.

Apply the idea

From part (a), we labeled the opposite and hypotenuse sides. The adjacent side is labeled x in the diagram. In order to find \cos \theta, we need to find the value of x and then write the trigonometric ratio.

\displaystyle a^2+b^2\displaystyle =\displaystyle c^2Write the formula
\displaystyle x^2+4^2\displaystyle =\displaystyle 5^2Substitute b=4,\,c=5
\displaystyle x^2+16\displaystyle =\displaystyle 25Simplify
\displaystyle x^2\displaystyle =\displaystyle 9Subtract 16 from both sides
\displaystyle x\displaystyle =\displaystyle 3Square root both sides

Now we want to write the trigonometric ratio for cosine, \dfrac{\text{Adjacent}}{\text{Hypotenuse}}

\cos\theta = \dfrac{3}{5}

Reflect and check

We could also realize that this a Pythagorean triple of 3, 4, \text{and } 5 at the start and then write the trigonometric ratio for \cos \theta.

c

Find the value of \tan \theta.

Worked Solution
Create a strategy

Use the tangent ratio.

Apply the idea

From part (a), we labeled the opposite and hypotenuse sides. From part (b) we labeled and found the value of the adjacent side. We can use these values to write the trigonometric ratio for tangent, \dfrac{\text{Opposite}}{\text{Adjacent}}:

\tan\theta = \dfrac{4}{3}

Idea summary
This image shows the relationship of the sine, cosine, and tangent ratios. Ask your teacher for more information.

\sin \theta =\dfrac{\text{Opposite }}{\text{Hypotenuse }}=\dfrac{\text{O}}{\text{H}} \qquad \cos \theta =\dfrac{\text{Adjacent }}{\text{Hypotenuse }}=\dfrac{\text{A}}{\text{H}} \qquad\tan \theta =\dfrac{\text{Opposite }}{\text{Adjacent }}=\dfrac{\text{O}}{\text{A}}

The mnemonic of SOH CAH TOA is helpful to remember the sides that apply to the different ratios of sine, cosine and tangent.

Find a side

For a right-angled triangle, given an angle and a side length, it is possible to use trigonometric ratios to find the length of an unknown side.

Using the reference angle, label the sides opposite, adjacent and hypotenuse. Choose which ratio to use and solve the equation for the unknown.

Examples

Example 2

Find the value of f, correct to two decimal places.

Right angled triangle with an angle of 25 degrees, adjacent side of 11 millimetres, and an unknown opposite side of f.
Worked Solution
Create a strategy

Use the tangent ratio.

Apply the idea

With respect to the angle of 25\degree, the side of length 11 is adjacent and the side of length f is opposite, so we will use the tangent ratio.

\displaystyle \tan \left(\theta\right)\displaystyle =\displaystyle \dfrac{\text{Opposite }}{\text{Adjacent}}Use the tangent ratio
\displaystyle \tan 25 \degree\displaystyle =\displaystyle \dfrac{f}{11}Substitute the values and f
\displaystyle 11 \times \tan 25 \degree\displaystyle =\displaystyle \dfrac{f}{11}\times11Multiply both sides by 11
\displaystyle f\displaystyle =\displaystyle 5.13 \text{ mm}Evaluate
Idea summary

We can use the trigonometric ratios to find an unknown side length of a right angled triangle.

Once we set up our equation with a pronumeral representing the unknown side length, we can use inverse operations to make the pronumeral the subject of the equation.

Then we can evaluate the expression in our calculators to find the side length.

Find an angle

Inverse trigonometric ratios are used to find unknown interior angles in right-angled triangles, providing at least two of the triangle's side-lengths are known.

A right angled triangle A B C with angle theta at A and right angle at B. Ask your teacher for more information.

Consider the following right-angled triangle with unknown interior angle \theta: \begin{aligned} \theta&=\sin^{-1}\left(\dfrac{\text{Opposite }}{\text{Hypotenuse }}\right)=\dfrac{\text{O}}{\text{H}} \\ \theta &=\cos^{-1}\left(\dfrac{\text{Adjacent }}{\text{Hypotenuse }}\right)=\dfrac{\text{A}}{\text{H}} \\ \theta &=\tan^{-1}\left(\dfrac{\text{Opposite }}{\text{Adjacent }}\right)=\dfrac{\text{O}}{\text{A}} \end{aligned}

Examples

Example 3

If \cos \theta=0.256, find \theta. Round your answer to two decimal places.

Worked Solution
Create a strategy

Use the inverse cosine function.

Apply the idea
\displaystyle \theta\displaystyle =\displaystyle \cos^{-1}\left(0.256\right)Apply inverse cosine both sides
\displaystyle \theta\displaystyle =\displaystyle 75.17 \degreeEvaluate using a calculator
Reflect and check

We need to use a calculator here to evaluate cosine inverse. Make sure your calculator is in degree mode when working with inverse trigonometric ratios.

Example 4

Find the value of x to the nearest degree.

Right angled triangle A B C with angle x at B, right angle at A, side A B has length of 7, and side B C has length of 25.
Worked Solution
Create a strategy

Use the cosine ratio to set up the equation then use the inverse ratio to find the angle.

Apply the idea

With respect to the angle of x, the length of the adjacent side is 7, and the length of the hypotenuse side is 25, so we can use the cosine ratio.

\displaystyle \cos x\displaystyle =\displaystyle \frac{\text{Adjacent }}{\text{Hypotenuse}}Use the cosine ratio
\displaystyle \cos x\displaystyle =\displaystyle \frac{7}{25}Substitute the values
\displaystyle x\displaystyle =\displaystyle \cos^{-1}\left(\frac{7}{25}\right)Apply the inverse ratio
\displaystyle \approx\displaystyle 74 \degreeEvaluate and round
Reflect and check

After identifying which sides we were given, we chose the inverse trigonometric ratio that matched those sides. We then solved the equation to find our unknown angle size.

Example 5

Consider the given figure.

A right angled triangle with sides of 25 and 30. Ask your teacher for more information.
a

Find the unknown angle x, correct to two decimal places.

Worked Solution
Create a strategy

Use the tangent ratio to set up the equation then use the inverse ratio to find the angle.

Apply the idea

With respect to the angle of x, the length of the opposite side is 15, and the length of the adjacent side is 30, so we can use the tangent ratio.

\displaystyle \tan \left(\theta \right)\displaystyle =\displaystyle \frac{\text{Opposite }}{\text{Adjacent}}Use the tangent ratio
\displaystyle \tan x\displaystyle =\displaystyle \dfrac{15}{30}Substitute the values and x
\displaystyle x\displaystyle =\displaystyle \tan^{-1}\left(\dfrac{15}{30}\right)Apply the inverse ratio
\displaystyle =\displaystyle 26.57 \degreeEvaluate and round
b

Find y, correct to two decimal places.

Worked Solution
Create a strategy

Use the tangent ratio to set up the equation then use the inverse ratio to find the angle.

Apply the idea

With respect to the angle of y, the length of the opposite side is 30, and the length of the adjacent side is 25, so we can use the tangent ratio.

\displaystyle \tan \left(\theta \right)\displaystyle =\displaystyle \frac{\text{Opposite }}{\text{Adjacent}}Use the tangent ratio
\displaystyle \tan y\displaystyle =\displaystyle \dfrac{30}{25}Substitute the values and y
\displaystyle y\displaystyle =\displaystyle \tan^{-1}\left(\dfrac{30}{25}\right)Apply the inverse ratio
\displaystyle =\displaystyle 50.19 \degreeEvaluate and round
c

Find z correct to two decimal places.

Worked Solution
Create a strategy

Use the fact that the total sum of the angles of a triangle is equal to 180\degree.

Apply the idea

We found in part (a) the angle, x=26.57\degree and in part (b), the angle, y=50.19\degree.

\displaystyle 180\displaystyle =\displaystyle 90+x+y+zAdd all the angles and equate to 180
\displaystyle 180\displaystyle =\displaystyle 90+26.57+50.19+zSubstitute the angles
\displaystyle 180\displaystyle =\displaystyle 166.76 + zSimplify right side
\displaystyle 180-166.76\displaystyle =\displaystyle 166.76 + z - 166.76Subtract 166.76 from both sides
\displaystyle z\displaystyle =\displaystyle 13.24\degreeEvaluate
Idea summary
A right angled triangle A B C with angle theta at A and right angle at B. Ask your teacher for more information.

Consider the following right-angled triangle with unknown interior angle \theta: \begin{aligned} \theta&=\sin^{-1}\left(\dfrac{\text{Opposite }}{\text{Hypotenuse }}\right)=\dfrac{\text{O}}{\text{H}} \\ \theta &=\cos^{-1}\left(\dfrac{\text{Adjacent }}{\text{Hypotenuse }}\right)=\dfrac{\text{A}}{\text{H}} \\ \theta &=\tan^{-1}\left(\dfrac{\text{Opposite }}{\text{Adjacent }}\right)=\dfrac{\text{O}}{\text{A}} \end{aligned}

Outcomes

U2.AoS4.2

Pythagoras’ theorem and the trigonometric ratios (sine, cosine and tangent) and their application including angles of elevation and depression and three figure bearings

U2.AoS4.9

solve practical problems involving right-angled triangles in the dimensions including the use of angles of elevation and depression, Pythagoras’ theorem trigonometric ratios sine, cosine and tangent and the use of three-figure (true) bearings in navigation

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