Year 10 Maths Advanced Trigonometry

Trigonometric Ratios

Q: What are the three trigonometric ratios?

Sine is opposite over hypotenuse, cosine is adjacent over hypotenuse and tangent is opposite over adjacent. The memory aid is SOH CAH TOA.

Q: Why does sin squared plus cos squared equal 1?

In a right-angled triangle, opposite squared plus adjacent squared equals hypotenuse squared. Dividing every term by hypotenuse squared gives sin squared theta plus cos squared theta equals 1.

Q: What is the relationship between sin and cos of complementary angles?

The sine of an angle equals the cosine of its complement: sin theta equals cos of 90 degrees minus theta. The side opposite one acute angle is adjacent to the other.

20 practice questions 1 video lesson Theory + worked examples

The trigonometric ratios compare the sides of a right-angled triangle for a chosen angle \(\theta\): \(\sin\theta=\dfrac{\text{opp}}{\text{hyp}}\), \(\cos\theta=\dfrac{\text{adj}}{\text{hyp}}\) and \(\tan\theta=\dfrac{\text{opp}}{\text{adj}}\). In NSW Year 10 Advanced maths you write these ratios exactly, as fractions or surds, and use one ratio to find the other two (NESA outcome MA5‑TRG‑C‑01).

You will find a missing side before writing a ratio, turn a decimal such as \(\tan\theta=0.75\) into a triangle with sides \(3\), \(4\) and \(5\). You will connect the gradient of a line to \(\tan\theta\), compare ratios without a calculator, and use \(\sin^2\theta+\cos^2\theta=1\) to test whether a pair of values is possible. Harder questions use algebraic sides and short proofs. It is part of Year 10 Advanced Trigonometry.

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Theory

The trigonometric ratios compare the sides of a right-angled triangle: \(\sin\theta = \dfrac{O}{H}\), \(\cos\theta = \dfrac{A}{H}\) and \(\tan\theta = \dfrac{O}{A}\). This Year 10 Advanced maths page covers exact ratios as fractions and surds, finding all three ratios from one, complementary angles, the identity \(\sin^2\theta + \cos^2\theta = 1\), and problems with perimeters and gradients.

In a right-angled triangle, the sides are named from an acute angle \(\theta\): the hypotenuse (opposite the right angle, the longest side), the opposite side (facing \(\theta\)) and the adjacent side (next to \(\theta\), not the hypotenuse).

The ratios \(\sin\theta\), \(\cos\theta\) and \(\tan\theta\) (SOH CAH TOA) depend only on the angle, not on the size of the triangle, because all right-angled triangles with angle \(\theta\) are similar. So a ratio such as \(\tan\theta = \dfrac{7}{9}\) tells you the shape: the sides could be \(7\) and \(9\), or \(7k\) and \(9k\).

An exact value is a fraction or a surd, not a rounded decimal. Give surd answers with a rational denominator.

The two acute angles of a right-angled triangle are complementary (they add to \(90^\circ\)), and the side opposite one is adjacent to the other.

The two acute angles add to \(90^\circ\). Opposite and adjacent swap between them, so \(\sin\theta = \cos(90^\circ - \theta)\) and \(\tan(90^\circ - \theta) = \dfrac{1}{\tan\theta}\).

With hypotenuse \(1\), the sides are \(\cos\theta\) and \(\sin\theta\). Pythagoras' theorem then gives \(\sin^2\theta + \cos^2\theta = 1\).

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

\tan θ = \frac{\sin θ}{\cos θ}

Ratio	Definition	Facts for acute \(\theta\)
\(\sin\theta\)	\(\dfrac{\text{opposite}}{\text{hypotenuse}}\)	between \(0\) and \(1\); increases as \(\theta\) increases
\(\cos\theta\)	\(\dfrac{\text{adjacent}}{\text{hypotenuse}}\)	between \(0\) and \(1\); decreases as \(\theta\) increases
\(\tan\theta\)	\(\dfrac{\text{opposite}}{\text{adjacent}}\)	any positive value; equals \(1\) at \(45^\circ\)

Identities. \(\sin^2\theta + \cos^2\theta = 1\) (divide \(O^2 + A^2 = H^2\) by \(H^2\)). Complementary angles: \(\sin\theta = \cos(90^\circ - \theta)\), \(\cos\theta = \sin(90^\circ - \theta)\). Dividing \(O^2 + A^2 = H^2\) by other squares gives further identities.

Gradient and angle. A line with gradient \(m\) makes an angle \(\theta\) with the positive \(x\)-axis where \(\tan\theta = m\) (rise over run).

How to find exact ratios

Label the hypotenuse, opposite and adjacent sides from the angle you are using.
Find any missing side with Pythagoras' theorem; simplify surds.
Write each ratio as a fraction in simplest form; rationalise surd denominators.

Other questions

Given one ratio: draw a triangle with those two sides (write a decimal as a fraction first, \(1.25 = \dfrac{5}{4}\)), then find the third side.
Given a ratio and a side: set up the ratio as an equation, or scale the triangle by \(k\). Use \(k\) for a perimeter too.
Composite figures: find a smaller right-angled triangle inside a square, rectangle or larger triangle (midpoints, perpendiculars, equal sides).
Algebraic sides: use Pythagoras' theorem to form a quadratic, keep the positive solution, then write the ratio.
Is it possible? Test \(\sin^2\theta + \cos^2\theta = 1\); also \(\sin\theta\) and \(\cos\theta\) must be less than \(1\).
Ordering without a calculator: use \(\cos\theta = \sin(90^\circ - \theta)\), that \(\sin\) increases as \(\theta\) increases, and that \(\tan\theta \gt 1\) once \(\theta \gt 45^\circ\).
Proofs: start from \(O^2 + A^2 = H^2\) or write a side as \(H\sin\theta\) or \(H\cos\theta\), then substitute. Squaring \(\sin\theta + \cos\theta\) brings in \(\sin^2\theta + \cos^2\theta\).

Example 1 — From one ratio to the others

For an acute angle \(\theta\), \(\cos\theta = \dfrac{20}{29}\). Find the exact values of
(i) \(\sin\theta\)
(ii) \(\tan\theta\)
(iii) \(\tan(90^\circ - \theta)\).

Solution

Draw a right-angled triangle with adjacent side \(20\) and hypotenuse \(29\). Find the opposite side:

\(\text{opp}^2\)	\(=\)	\(29^2 - 20^2 = 441\)
\(\text{opp}\)	\(=\)	\(21\)

\(\text{(i) } \sin\theta\)	\(=\)	\(\boldsymbol{\dfrac{21}{29}}\)
\(\text{(ii) } \tan\theta\)	\(=\)	\(\boldsymbol{\dfrac{21}{20}}\)
\(\text{(iii) } \tan(90^\circ - \theta)\)	\(=\)	\(\boldsymbol{\dfrac{20}{21}}\)

\sin θ = \frac{21}{29}

For the other acute angle, \(90^\circ - \theta\), the \(21\) side is adjacent and the \(20\) side is opposite.

Example 2 — A surd hypotenuse

A right-angled triangle has shorter sides \(2\) cm and \(5\) cm. The angle \(\theta\) is opposite the \(2\) cm side. Find \(\sin\theta\) and \(\cos\theta\) exactly, with rational denominators.

Solution

\(\text{hyp}^2\)	\(=\)	\(2^2 + 5^2 = 29\)
\(\text{hyp}\)	\(=\)	\(\sqrt{29}\)

\(\sin\theta\)	\(=\)	\(\dfrac{2}{\sqrt{29}} \times \dfrac{\sqrt{29}}{\sqrt{29}} = \boldsymbol{\dfrac{2\sqrt{29}}{29}}\)
\(\cos\theta\)	\(=\)	\(\dfrac{5}{\sqrt{29}} \times \dfrac{\sqrt{29}}{\sqrt{29}} = \boldsymbol{\dfrac{5\sqrt{29}}{29}}\)

\sin θ = \frac{2 \sqrt{29}}{29}

\(29\) has no square factor, so \(\sqrt{29}\) is already in simplest form.

Example 3 — A ratio and one side

A ramp is \(8.2\) m long and makes an angle \(\theta\) with the horizontal, where \(\sin\theta = \dfrac{9}{41}\). Find the rise and the horizontal run of the ramp.

Solution

The ramp is the hypotenuse and the rise is opposite \(\theta\):

\(\dfrac{\text{rise}}{8.2}\)	\(=\)	\(\dfrac{9}{41}\)
\(\text{rise}\)	\(=\)	\(8.2 \times \dfrac{9}{41} = \boldsymbol{1.8}\text{ m}\)

A \(9\), \(40\), \(41\) triangle gives \(\cos\theta = \dfrac{40}{41}\), since \(41^2 - 9^2 = 40^2\):

\(\text{run}\)

\(=\)

\(8.2 \times \dfrac{40}{41} = \boldsymbol{8}\text{ m}\)

run = 8.2 \times \frac{40}{41} = 8

Check: \(1.8^2 + 8^2 = 3.24 + 64 = 67.24 = 8.2^2\) ✓

Example 4 — A ratio and the perimeter

A right-angled triangle has a perimeter of \(120\) cm. For one acute angle \(\theta\), \(\tan\theta = \dfrac{8}{15}\). Find the three sides, then check that \(\sin^2\theta + \cos^2\theta = 1\).

Solution

The opposite and adjacent sides are in the ratio \(8 : 15\). Call them \(8k\) and \(15k\):

\(\text{hyp}\)	\(=\)	\(\sqrt{(8k)^2 + (15k)^2} = \sqrt{289k^2} = 17k\)
\(8k + 15k + 17k\)	\(=\)	\(120\)
\(40k\)	\(=\)	\(120, \ \ k = 3\)

The sides are \(\boldsymbol{24}\) cm, \(\boldsymbol{45}\) cm and \(\boldsymbol{51}\) cm.

\(\sin^2\theta + \cos^2\theta\)	\(=\)	\(\left(\dfrac{24}{51}\right)^2 + \left(\dfrac{45}{51}\right)^2\)
\(\)	\(=\)	\(\dfrac{576 + 2025}{2601} = 1 \ ✓\)

\sin^{2} θ + \cos^{2} θ = 1

Common pitfalls

Label from the angle you are using. The opposite and adjacent sides change when you change the angle; the hypotenuse never does.

Hypotenuse on the bottom. \(\sin\) and \(\cos\) always have the hypotenuse as the denominator, so they are less than \(1\).

\(\sin^2\theta\) means \((\sin\theta)^2\). It is not \(\sin(\theta^2)\).

Ratios are not lengths. \(\tan\theta = \dfrac{2}{5}\) does not mean the sides are \(2\) and \(5\); they could be \(4\) and \(10\).

Rationalise surds. Write \(\dfrac{3}{\sqrt{7}}\) as \(\dfrac{3\sqrt{7}}{7}\) when an exact answer with a rational denominator is asked for.

Frequently asked questions

What are the three trigonometric ratios?

\(\sin\theta = \dfrac{O}{H}\), \(\cos\theta = \dfrac{A}{H}\) and \(\tan\theta = \dfrac{O}{A}\). The memory aid is SOH CAH TOA.

How do you find sin and cos if you know tan?

Draw a right-angled triangle with the opposite and adjacent sides from the tangent fraction, find the hypotenuse with Pythagoras' theorem, then read off sine and cosine.

Why does sin squared plus cos squared equal 1?

In a right-angled triangle \(O^2 + A^2 = H^2\). Dividing every term by \(H^2\) gives \(\sin^2\theta + \cos^2\theta = 1\).

What is the relationship between sin and cos of complementary angles?

The sine of an angle equals the cosine of its complement: \(\sin\theta = \cos(90^\circ - \theta)\). The side opposite one acute angle is adjacent to the other.

Do the trigonometric ratios depend on the size of the triangle?

No. All right-angled triangles with the same acute angle are similar, so their sides are in the same ratios. Only the angle matters.

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Trigonometric Ratios

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How to find exact ratios

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