Finding an angle | Year 9 | Ascent Library

01 / LEARN

Understand the idea.

Work backwards from a ratio

When two sides are known, write their trigonometric ratio first. An inverse function then finds the acute angle producing that ratio. Use inverse sine for opposite and hypotenuse, inverse cosine for adjacent and hypotenuse, or inverse tangent for the legs.

Enter the whole fraction inside the inverse function. Here ${sin}^{- 1}$ , ${cos}^{- 1}$ and ${tan}^{- 1}$ mean inverse functions, not reciprocals. For example, ${tan}^{- 1} (r)$ finds an angle; $\frac{1}{tan (r)}$ performs a different calculation.

θ = {tan}^{- 1} (\frac{o}{a})

The answer must fit the triangle

Use degree mode and check that the result is between $0^{\circ}$ and $90^{\circ}$ . In a non-degenerate right triangle, opposite-to-hypotenuse and adjacent-to-hypotenuse ratios must lie strictly between zero and one. Larger values indicate a labelling or data error.

If opposite is longer than adjacent, the chosen angle exceeds $45^{\circ}$ ; if it is shorter, the angle is smaller. The other acute angle is $90^{\circ} - θ$ . These checks help reveal using the correct ratio for the wrong angle.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Opposite and adjacent

Opposite = 9 cm and adjacent = 12 cm. Find θ.

Worked solution

1
Write the tangent ratio.
$tan θ = \frac{9}{12} = 0.75$
The two known sides are opposite and adjacent.
2
Apply inverse tangent in degree mode.
$\begin{aligned} θ & = {tan}^{- 1} (0.75) \\ = 36.869897^{\dots \circ} \end{aligned}$
Inverse tangent returns the acute angle whose opposite-to-adjacent ratio is three quarters.
3
Round and check the angle.
$θ \approx {36.87}^{\circ} < 45^{\circ}$
The opposite side is shorter than the adjacent side, so an angle below forty-five degrees is sensible.

Answer

36.87°

EXAMPLE 02

Adjacent and hypotenuse

Adjacent = 7 m and hypotenuse = 10 m. Find θ.

Worked solution

1
Write the cosine ratio.
$cos θ = \frac{7}{10} = 0.7$
The adjacent side and hypotenuse select cosine.
2
Apply inverse cosine in degree mode.
$\begin{aligned} θ & = {cos}^{- 1} (0.7) \\ = 45.572995^{\dots \circ} \end{aligned}$
Inverse cosine returns the angle with this adjacent-to-hypotenuse ratio.
3
Round to two decimal places.
$θ \approx {45.57}^{\circ}$
The next decimal digit is two, and the result lies within the acute-angle range.

Answer

45.57°

03 / YOUR TURN

Try it yourself.

Check your understanding

Opposite and adjacent are equal. Find θ.

Show answer & reasoning

45°

Equal opposite and adjacent lengths give $tan θ = \frac{o}{a} = 1$ . Therefore $θ = {tan}^{- 1} (1) = 45^{\circ}$ in this acute-angle setting.

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

Find $θ$ if $o = 5$ cm and $h = 13$ cm. Give two decimal places.
2 marks
Find $θ$ if $a = 12$ cm and $h = 13$ cm. Give two decimal places.
2 marks
Find $θ$ if $o = 15$ m and $a = 8$ m. Give two decimal places.
2 marks
An acute angle has sine 0.8. Find it to two decimal places.
2 marks
For hypotenuse 13 and opposite 5, a student enters ${sin}^{- 1} (\frac{13}{5})$ . Explain the error and correct it.
3 marks
One acute angle in a right triangle is ${34.7}^{\circ}$ . Find the other.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find $θ$ if $o = 5$ cm and $h = 13$ cm. Give two decimal places.
2 marks
$θ = {sin}^{- 1} (\frac{5}{13})$ .
${22.62}^{\circ}$
Find $θ$ if $a = 12$ cm and $h = 13$ cm. Give two decimal places.
2 marks
$θ = {cos}^{- 1} (\frac{12}{13})$ .
This agrees with the matching 5–12–13 triangle.
${22.62}^{\circ}$
Find $θ$ if $o = 15$ m and $a = 8$ m. Give two decimal places.
2 marks
$θ = {tan}^{- 1} (\frac{15}{8})$ .
Opposite exceeds adjacent, so $θ$ exceeds $45^{\circ}$ .
${61.93}^{\circ}$
An acute angle has sine 0.8. Find it to two decimal places.
2 marks
$θ = {sin}^{- 1} (0.8)$ .
${53.13}^{\circ}$
For hypotenuse 13 and opposite 5, a student enters ${sin}^{- 1} (\frac{13}{5})$ . Explain the error and correct it.
3 marks
The side ratio was reversed.
$\frac{13}{5}$ exceeds 1 and cannot be a sine value.
Use ${sin}^{- 1} (\frac{5}{13})$
One acute angle in a right triangle is ${34.7}^{\circ}$ . Find the other.
2 marks
$90^{\circ} - {34.7}^{\circ} = {55.3}^{\circ}$ .
${55.3}^{\circ}$

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Curriculum reference & lesson scope

Australian Curriculum v9 and SA Mathematics

Year 9 similarity, Pythagoras and right-triangle trigonometric ratios, with prerequisite revision. Non-right-triangle sine/cosine rules are not required.

Official curriculum source ↗

In this lesson

Understand the idea.

Work backwards from a ratio

The answer must fit the triangle

See the method in action.

Opposite and adjacent

Write the tangent ratio.

Apply inverse tangent in degree mode.

Round and check the angle.

Adjacent and hypotenuse

Write the cosine ratio.

Apply inverse cosine in degree mode.

Round to two decimal places.

Try it yourself.

Check your understanding

Keep practising