Year 9 Maths Advanced Trigonometry

Trig Ratios

20 practice questions 2 video lessons Theory + worked examples

The trigonometric ratios compare two sides in a right-angled triangle: \(\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}\) and \(\tan\theta = \frac{\text{opposite}}{\text{adjacent}}\). This NSW Year 9 Advanced maths subtopic introduces them (NESA outcome MA5‑TRG‑C‑01).

You will label sides from an angle, write each ratio as a fraction, and see that enlarging a triangle does not change the ratios. You will use a calculator with degrees and minutes, find one ratio from another with Pythagoras' theorem, and find exact ratios in an isosceles triangle and from a point on the number plane. Harder questions prove \(\sin^2\theta + \cos^2\theta = 1\). It is part of Year 9 Advanced Trigonometry.

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Practice questions

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Trig Ratios - Video - Basic Trigonometry 1 Watch
Trig Ratios - Video - Basic trigonometry 2 Watch

Theory

The trigonometric ratios \(\sin\theta\), \(\cos\theta\) and \(\tan\theta\) compare pairs of sides of a right-angled triangle. This Year 9 Advanced maths page covers naming sides, why the ratios depend only on the angle, exact values from one ratio, calculator work in degrees and minutes, and harder problems with scale factors and equal angles.

In a right-angled triangle, name the sides from an acute angle \(\theta\). The hypotenuse is opposite the right angle and is the longest side. The opposite side faces \(\theta\). The adjacent side runs from \(\theta\) to the right angle.

The three trigonometric ratios are \(\sin\theta = \dfrac{O}{H}\), \(\cos\theta = \dfrac{A}{H}\) and \(\tan\theta = \dfrac{O}{A}\) (SOH CAH TOA).

Every right-angled triangle with the same angle \(\theta\) is similar, so its sides are in the same proportion. That is why each ratio depends only on the angle, not on the size of the triangle.

Sides are often named with small letters: side \(a\) is opposite vertex \(A\). In \(\triangle XYZ\) right-angled at \(Y\), \(\sin X = \dfrac{x}{y}\) because \(x\) faces \(X\) and \(y\) is the hypotenuse.

The two acute angles add to \(90^\circ\). The side opposite one is adjacent to the other, so \(\sin A = \cos B\) when \(A + B = 90^\circ\).

A small letter names the side opposite the vertex with the same capital letter. The hypotenuse \(c\) is opposite the right angle \(C\).

Enlarging a right-angled triangle multiplies every side by \(k\), so each trig ratio of \(\theta\) stays the same.

\[\sin\theta = \dfrac{O}{H} \qquad \cos\theta = \dfrac{A}{H} \qquad \tan\theta = \dfrac{O}{A}\]

\sin θ = \frac{O}{H}, \cos θ = \frac{A}{H}, \tan θ = \frac{O}{A}

Two results that follow from the definitions and Pythagoras' theorem:

\[\tan\theta = \dfrac{\sin\theta}{\cos\theta} \qquad \sin^2\theta + \cos^2\theta = 1\]

\tan θ = \frac{\sin θ}{\cos θ}, \sin^{2} θ + \cos^{2} θ = 1

Proving results like these. Call the sides opposite, adjacent and hypotenuse \(a\), \(b\) and \(c\). Write each ratio as a fraction in \(a\), \(b\) and \(c\), simplify, and use Pythagoras' theorem \(a^2 + b^2 = c^2\) where squares appear.

Complementary angles. \(\sin\theta = \cos(90^\circ - \theta)\) and \(\cos\theta = \sin(90^\circ - \theta)\).

Degrees and minutes. \(1^\circ = 60'\). To find an angle, use \(\sin^{-1}\), \(\cos^{-1}\) or \(\tan^{-1}\) and convert the decimal part of the degree to minutes.

How to write a trig ratio

Mark the angle and the right angle. Label H first (opposite the right angle), then O and A.
Write the ratio as a fraction of side lengths and simplify it.
Check: sine and cosine are less than \(1\) because the hypotenuse is the longest side; tangent can be any positive value.

From one ratio to the others

Sketch a triangle using the numerator and denominator as side lengths.
Find the third side with Pythagoras' theorem.
Write the other ratios exactly. Combine them as fractions if asked (for example \(\sin\theta + \cos\theta\)).
A real length (a hypotenuse, a side) gives the scale factor; multiply every side by it for perimeters and areas.

Harder diagrams

Isosceles triangle: the line from the apex to the midpoint of the base is perpendicular to it, making two right-angled triangles.
Altitude inside a right-angled triangle: each small angle at the right angle equals the opposite acute angle of the big triangle.
A point \((x, y)\) joined to the origin: the run is \(x\), the rise is \(y\) and the hypotenuse is \(\sqrt{x^2 + y^2}\).

Example 1 — One ratio gives the others

For an acute angle \(\theta\), \(\cos\theta = \dfrac{33}{65}\). Find the exact values of \(\sin\theta\), \(\tan\theta\) and \(\sin\theta - \cos\theta\).

Solution

Sketch a triangle with adjacent side \(33\) and hypotenuse \(65\), then find the opposite side with Pythagoras' theorem.

\(x^2\)	\(=\)	\(65^2 - 33^2 = 4225 - 1089 = 3136\)
\(x\)	\(=\)	\(56\)
\(\sin\theta\)	\(=\)	\(\boldsymbol{\dfrac{56}{65}}\)
\(\tan\theta\)	\(=\)	\(\boldsymbol{\dfrac{56}{33}}\)
\(\sin\theta - \cos\theta\)	\(=\)	\(\dfrac{56}{65} - \dfrac{33}{65} = \boldsymbol{\dfrac{23}{65}}\)

\sin θ = \frac{56}{65}, \tan θ = \frac{56}{33}

Exact means leave the answers as fractions, not rounded decimals.

Example 2 — Calculator, degrees and minutes

(i) \(\sin\theta = 0.4\) for an acute angle \(\theta\). Find \(\theta\), correct to the nearest minute.
(ii) Evaluate \(15\cos 52^\circ 40'\), correct to two decimal places.

Solution

(i) Use the inverse key: \(\sin^{-1}\) gives the angle whose sine is \(0.4\).

\(\theta\)	\(=\)	\(\sin^{-1} 0.4\)
\(\)	\(=\)	\(23.5781\ldots^\circ\)
\(\)	\(=\)	\(23^\circ\ 34.69'\)
\(\)	\(\approx\)	\(\mathbf{23^\circ 35'}\)

θ = \sin^{- 1} 0.4 \approx 23 ° 35'

The decimal part \(0.578\ldots^\circ\) times \(60\) gives the minutes. The degrees-minutes key does this for you.

(ii)

\(15\cos 52^\circ 40'\)	\(=\)	\(9.0967\ldots\)
\(\)	\(\approx\)	\(\mathbf{9.10}\)

15 \cos 52 ° 40' \approx 9.10

Enter \(52^\circ 40'\) with the degrees-minutes key, not as \(52.40^\circ\).

Example 3 — Ratio plus one length

In a right-angled triangle, \(\tan\theta = \dfrac{48}{55}\) and the hypotenuse is \(146\) cm. Find the perimeter and the area of the triangle.

Solution

The sides are in the ratio opposite : adjacent \(= 48 : 55\). Find the hypotenuse of the \(48\)-\(55\) triangle, then the scale factor.

\(\sqrt{48^2 + 55^2}\)	\(=\)	\(\sqrt{5329} = 73\)
\(k\)	\(=\)	\(146 \div 73 = 2\)
\(\text{Sides}\)	\(=\)	\(96,\ 110,\ 146\text{ cm}\)
\(\text{Perimeter}\)	\(=\)	\(96 + 110 + 146 = \mathbf{352}\text{ cm}\)
\(\text{Area}\)	\(=\)	\(\dfrac{1}{2} \times 96 \times 110 = \mathbf{5280}\text{ cm}^2\)

P = 352, A = 5280

A ratio fixes the shape, not the size. One real length fixes the scale factor.

Example 4 — Finding an equal angle

In \(\triangle ABC\), \(\angle ACB = 90^\circ\), \(AC = 12\) cm and \(BC = 16\) cm. \(CD\) is perpendicular to \(AB\). Find the exact value of \(\sin\angle BCD\).

Solution

\(\angle BCD\) sits in a triangle with no known sides. Show it equals an angle in \(\triangle ABC\) first.

\(\triangle BCD\) has a right angle at \(D\) and \(\triangle ABC\) has a right angle at \(C\). In each, the other two angles add to \(90^\circ\).

\(\angle BCD\)	\(=\)	\(90^\circ - \angle B\)
\(\angle A\)	\(=\)	\(90^\circ - \angle B\)
\(\therefore\ \angle BCD\)	\(=\)	\(\angle A\)
\(AB\)	\(=\)	\(\sqrt{12^2 + 16^2} = 20\)
\(\sin\angle BCD\)	\(=\)	\(\sin A = \dfrac{BC}{AB}\)
\(\)	\(=\)	\(\dfrac{16}{20} = \boldsymbol{\dfrac{4}{5}}\)

\sin ∠ B C D = \frac{4}{5}

Angles that add to \(90^\circ\) with the same angle are equal. This trick appears whenever a perpendicular is dropped inside a right-angled triangle.

Common pitfalls

Name the sides from the right angle. The hypotenuse never changes. Opposite and adjacent swap when you move to the other acute angle.

Minutes are not decimals. \(47^\circ 20'\) is \(47.333\ldots^\circ\), not \(47.2^\circ\). Use the degrees-minutes key.

Bigger triangle, same ratio. Enlarging a triangle by a scale factor does not change \(\sin\theta\), \(\cos\theta\) or \(\tan\theta\). Only the angle does.

Keep exact answers exact. If a question asks for an exact value, leave it as a fraction or surd such as \(\dfrac{56}{65}\), not \(0.86\).

Check the calculator mode. It must be in degrees. In radian mode every answer is wrong.

Frequently asked questions

What is SOH CAH TOA?

It is a memory aid for the three trig ratios: Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, and Tangent is Opposite over Adjacent.

Why are trig ratios the same for any size of triangle?

Right-angled triangles with the same acute angle are similar, so their sides are in the same proportion. Each ratio of two sides therefore depends only on the angle.

How do you find cos if you know sin?

Draw a right-angled triangle using the sine fraction as opposite over hypotenuse, find the adjacent side with Pythagoras' theorem, then write adjacent over hypotenuse. For sin = 16 over 65, the adjacent side is 63, so cos = 63 over 65.

How do you find an angle in degrees and minutes on a calculator?

Use the inverse key, such as shift then sin, to get the angle as a decimal. Then press the degrees-minutes-seconds key, or multiply the decimal part by 60 to get the minutes.

Is sin A equal to cos B in a right-angled triangle?

Yes, when A and B are the two acute angles. The side opposite A is adjacent to B, so sin A and cos B are the same fraction.

Can sin or cos be bigger than 1?

Not for an angle in a right-angled triangle. Sine and cosine divide a shorter side by the hypotenuse, which is always the longest side.

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Finding Unknown Side

Trig Ratios

📖 Theory

How to write a trig ratio

From one ratio to the others

Harder diagrams

Common pitfalls

Frequently asked questions

Theory