Sine/cosine/tangent | Year 11

01 / LEARN

Understand the idea.

Ratios are tied to the reference angle

Label opposite, adjacent and hypotenuse relative to the chosen acute angle. Sine uses opposite and hypotenuse; cosine uses adjacent and hypotenuse; tangent uses the two legs. Choose the ratio containing the given side and the unknown.

Similar right triangles explain why these ratios depend on angle rather than triangle size. Set the calculator to degrees for this chapter and write a ratio equation before calculating, especially when the unknown is in a denominator.

\begin{aligned} sin θ & = \frac{O}{H} \\ cos θ & = \frac{A}{H} \\ tan θ & = \frac{O}{A} \end{aligned}

Recover an angle from a ratio

An inverse trigonometric function recovers an angle from a ratio. In a right triangle the reference angle is acute, so the required value lies between $0^{\circ}$ and $90^{\circ}$ . The calculator must return degrees here.

The notation ${sin}^{- 1}$ means inverse sine in this context, not reciprocal sine. Keep unrounded ratios until the final answer. Check that a calculated leg is shorter than the hypotenuse and that the chosen ratio matches the labelled sides.

θ = {tan}^{- 1} (\frac{O}{A})

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

A right triangle has hypotenuse 12 m and an acute angle of 35°. Find the opposite side. Round to one decimal place.

Worked solution

1
Choose the ratio involving opposite and hypotenuse.
$sin 35^{\circ} = \frac{x}{12}$
The unknown is opposite the reference angle and twelve metres is the hypotenuse.
2
Isolate the length.
$x = 12 sin 35^{\circ}$
Multiply both sides by the hypotenuse length.
3
Calculate and round once.
$x \approx 6.8829 m \approx 6.9 m$
The question requests one decimal place; the result is shorter than the twelve-metre hypotenuse.

Answer

6.9 m

EXAMPLE 02

Worked example 2

A right triangle has opposite leg 7 cm and adjacent leg 11 cm relative to θ. Find θ to one decimal place.

Worked solution

1
Form the tangent ratio.
$tan θ = \frac{7}{11}$
The known lengths are the opposite and adjacent legs.
2
Apply inverse tangent.
$θ = {tan}^{- 1} (\frac{7}{11})$
Inverse tangent returns the acute angle having this leg ratio.
3
Calculate in degrees and round.
$θ \approx {32.4712}^{\circ} \approx {32.5}^{\circ}$
Rounding at the end preserves the requested one-decimal accuracy.

Answer

32.5°

03 / YOUR TURN

Try it yourself.

Check your understanding

A right triangle has opposite leg 3 and hypotenuse 5. Find sinθ.

Show answer & reasoning

0.6

Use opposite divided by hypotenuse: $sin θ = \frac{3}{5} = 0.6$ . No inverse function is needed because the question asks for a ratio rather than the angle.

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

A right triangle has opposite leg 8 and hypotenuse 17. Find $sin θ$ as a fraction.
2 marks
A right triangle has adjacent leg 12 and hypotenuse 13. Find $cos θ$ .
2 marks
A right triangle has opposite leg 9 and adjacent leg 4. Find $tan θ$ .
2 marks
Find the opposite leg when the adjacent leg is 14 m and $θ = 28^{\circ}$ . Round to one decimal place.
2 marks
Find an acute angle with opposite leg 5 cm and hypotenuse 9 cm. Round to one decimal place.
2 marks
A calculation gives $sin θ = 1.2$ for a right triangle. Can an acute angle satisfy this? Explain.
3 marks

Read the worked solutions

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

A right triangle has opposite leg 8 and hypotenuse 17. Find $sin θ$ as a fraction.
2 marks
Sine is opposite divided by hypotenuse.
$\frac{8}{17}$
A right triangle has adjacent leg 12 and hypotenuse 13. Find $cos θ$ .
2 marks
Cosine is adjacent divided by hypotenuse.
$\frac{12}{13}$
A right triangle has opposite leg 9 and adjacent leg 4. Find $tan θ$ .
2 marks
Tangent is opposite divided by adjacent.
$\frac{9}{4}$ or 2.25
Find the opposite leg when the adjacent leg is 14 m and $θ = 28^{\circ}$ . Round to one decimal place.
2 marks
$tan 28^{\circ} = \frac{x}{14}$ .
$x = 14 tan 28^{\circ}$ .
$x \approx 7.4 m$ .
7.4 m
Find an acute angle with opposite leg 5 cm and hypotenuse 9 cm. Round to one decimal place.
2 marks
$sin θ = \frac{5}{9}$ .
$θ = {sin}^{- 1} (\frac{5}{9})$ .
$θ \approx {33.7}^{\circ}$ .
${33.7}^{\circ}$
A calculation gives $sin θ = 1.2$ for a right triangle. Can an acute angle satisfy this? Explain.
3 marks
The opposite side is shorter than the hypotenuse, so $0 < sin θ < 1$ .
A value 1.2 indicates an incorrect ratio or inconsistent data.
No.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 General Mathematics: practical trigonometry, including similarity, right triangles, triangle area, sine/cosine rules and contextual interpretation.

Official curriculum source ↗

In this lesson

Understand the idea.

Ratios are tied to the reference angle

Recover an angle from a ratio

See the method in action.

Worked example 1

Choose the ratio involving opposite and hypotenuse.

Isolate the length.

Calculate and round once.

Worked example 2

Form the tangent ratio.

Apply inverse tangent.

Calculate in degrees and round.

Try it yourself.

Check your understanding

Keep practising