Year 11 Methods (Unit 1 & 2) Trigonometric Functions

Reviewing Trigonometric Ratios

Q: What does SOH CAH TOA mean?

It codes the three ratios. Sine equals opposite over hypotenuse. Cosine equals adjacent over hypotenuse. Tangent equals opposite over adjacent.

Q: What is an angle of elevation?

It is the angle measured up from the horizontal to a line of sight. In a right triangle sketch it is the acute angle at the observer. An angle of depression is measured down from the horizontal.

Q: What are the exact ratios of the special angles?

From the special triangles: sine of thirty degrees is one half, cosine of thirty degrees is root three over two, tangent of forty five degrees is one, and tangent of sixty degrees is root three.

20 practice questions 1 video lesson Theory + worked examples

Review right-triangle trigonometry for Queensland Year 11 Mathematical Methods (QCAA). Using SOH CAH TOA, the sine, cosine and tangent ratios link an acute angle to the sides of a right-angled triangle.

You will learn to find an unknown side, use inverse trigonometry for an unknown angle, apply the exact ratios of thirty, forty-five and sixty degrees, and solve elevation and depression problems — practical skills underpinning the trigonometry unit in the QCAA course.

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Theory

In Year 11 Mathematical Methods (QCAA, Unit 1), Topic 4 opens by reviewing right-triangle trigonometry. The ratios SOH CAH TOA connect an acute angle to the opposite, adjacent and hypotenuse sides, letting you find an unknown side or angle, solve angle-of-elevation problems, and read the exact ratios of \(30^\circ\), \(45^\circ\) and \(60^\circ\) from the special triangles.

In a right-angled triangle, name the sides relative to a chosen acute angle \(\theta\): the hypotenuse is the longest side (opposite the right angle), the opposite is the side across from \(\theta\), and the adjacent is the remaining side next to \(\theta\).

The three ratios are remembered as SOH CAH TOA: \(\sin\theta=\dfrac{\text{opp}}{\text{hyp}}\), \(\cos\theta=\dfrac{\text{adj}}{\text{hyp}}\), \(\tan\theta=\dfrac{\text{opp}}{\text{adj}}\). To find an unknown side, substitute and rearrange; to find an unknown angle, use the inverse functions \(\sin^{-1}\), \(\cos^{-1}\), \(\tan^{-1}\).

The special triangles (the \(45\text{-}45\text{-}90\) with sides \(1,1,\sqrt2\) and the \(30\text{-}60\text{-}90\) with sides \(1,\sqrt3,2\)) give exact ratios such as \(\sin 30^\circ=\dfrac{1}{2}\) and \(\tan 60^\circ=\sqrt3\).

Choose the ratio by the sides involved. Use the one that links the angle to the side you know and the side you want — opposite and hypotenuse means sine, and so on.

The sides named relative to \(\theta\): opposite, adjacent and hypotenuse.

The special triangles: \(45\text{-}45\text{-}90\) (\(1,1,\sqrt2\)) and \(30\text{-}60\text{-}90\) (\(1,\sqrt3,2\)).

The three primary ratios (SOH CAH TOA):

\[\sin\theta=\dfrac{\text{opp}}{\text{hyp}},\quad \cos\theta=\dfrac{\text{adj}}{\text{hyp}},\quad \tan\theta=\dfrac{\text{opp}}{\text{adj}}\]

\sin θ = \frac{opp}{hyp}

Finding an unknown angle uses the inverse functions:

\[\theta=\tan^{-1}\!\left(\dfrac{\text{opp}}{\text{adj}}\right)\]

θ = \tan^{- 1} (\frac{opp}{adj})

Exact ratios from the special triangles:

\[\sin 30^\circ=\dfrac{1}{2},\ \ \cos 30^\circ=\dfrac{\sqrt3}{2},\ \ \tan 45^\circ=1,\ \ \tan 60^\circ=\sqrt3\]

\sin 30^{\circ} = \frac{1}{2}

Elevation and depression are both measured from the horizontal; in a right-triangle sketch they sit at the observer as the acute angle \(\theta\).

How to solve a right-triangle problem

Label the sides relative to the angle: opposite, adjacent, hypotenuse.
Choose the ratio (SOH CAH TOA) that links the known side(s) to the unknown, and write the equation.
Solve: rearrange for an unknown side, or apply \(\sin^{-1}/\cos^{-1}/\tan^{-1}\) for an unknown angle; keep surds exact where asked.

Example 1 — Finding a side with cosine

In a right triangle the hypotenuse is \(10\) cm and one acute angle is \(60^\circ\). Find the length \(x\) of the side adjacent to that angle.

Solution

The adjacent and hypotenuse are involved, so use \(\cos\theta=\dfrac{\text{adj}}{\text{hyp}}\):

\(\cos 60^\circ\)

\(=\)

\(\dfrac{x}{10}\)

Multiply both sides by \(10\) and use \(\cos 60^\circ=\dfrac{1}{2}\):

\(x\)	\(=\)	\(10\cos 60^\circ\)
	\(=\)	\(10\times\dfrac{1}{2}\)
	\(=\)	\(5\)

The adjacent side is \(x=5\) cm.

x = 5

Example 2 — Finding an angle

A right triangle has an opposite side of \(8\) and an adjacent side of \(8\) relative to the angle \(\theta\). Find \(\theta\).

Solution

Opposite and adjacent are involved, so use \(\tan\theta=\dfrac{\text{opp}}{\text{adj}}\):

\(\tan\theta\)	\(=\)	\(\dfrac{8}{8}\)
	\(=\)	\(1\)

Take the inverse tangent:

\(\theta\)	\(=\)	\(\tan^{-1}(1)\)
	\(=\)	\(45^\circ\)

The angle is \(\theta=45^\circ\).

θ = 45^{\circ}

Example 3 — Angle of elevation

From a point \(40\) m from the base of a tower, the angle of elevation to the top is \(30^\circ\). Find the exact height \(h\) of the tower.

Solution

The height is opposite the \(30^\circ\) angle and \(40\) m is adjacent, so use \(\tan\theta=\dfrac{\text{opp}}{\text{adj}}\):

\(\tan 30^\circ\)

\(=\)

\(\dfrac{h}{40}\)

Rearrange and use \(\tan 30^\circ=\dfrac{1}{\sqrt3}\):

\(h\)	\(=\)	\(40\tan 30^\circ\)
	\(=\)	\(\dfrac{40}{\sqrt3}\)
	\(=\)	\(\dfrac{40\sqrt3}{3}\)

The height is \(h=\dfrac{40\sqrt3}{3}\approx 23.1\) m.

h = \frac{40 \sqrt{3}}{3}

Example 4 — Exact sides from a special triangle

A right triangle has a \(30^\circ\) angle and the side opposite it measures \(7\) cm. Find the exact hypotenuse and adjacent side.

Solution

Opposite and hypotenuse use sine, \(\sin 30^\circ=\dfrac{1}{2}\):

\(\sin 30^\circ\)	\(=\)	\(\dfrac{7}{\text{hyp}}\)
\(\text{hyp}\)	\(=\)	\(\dfrac{7}{\sin 30^\circ}\)
	\(=\)	\(\dfrac{7}{1/2}\)
	\(=\)	\(14\)

Opposite and adjacent use tangent, \(\tan 30^\circ=\dfrac{1}{\sqrt3}\):

\(\tan 30^\circ\)	\(=\)	\(\dfrac{7}{\text{adj}}\)
\(\text{adj}\)	\(=\)	\(\dfrac{7}{\tan 30^\circ}\)
	\(=\)	\(7\sqrt3\)

The hypotenuse is \(14\) cm and the adjacent side is \(7\sqrt3\) cm.

hyp = 14, adj = 7 \sqrt{3}

Common pitfalls

Mixing up opposite and adjacent. These are named relative to the angle you are using; the opposite is across from the angle, the adjacent is beside it (and is not the hypotenuse).

Forgetting to use the inverse for an angle. If you know a ratio and want the angle, apply \(\sin^{-1}\), \(\cos^{-1}\) or \(\tan^{-1}\) — not the ratio itself.

Calculator in radian mode. These are degree problems; make sure the calculator is set to degrees, or \(\cos 60^\circ\) will not give \(0.5\).

Frequently asked questions

What does SOH CAH TOA mean?

It codes the three ratios: Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent.

How do you decide which ratio to use?

Look at which two sides the problem links to the angle. Opposite and hypotenuse means sine, adjacent and hypotenuse means cosine, opposite and adjacent means tangent.

How do you find an unknown angle in a right triangle?

Form the ratio from the two known sides, then apply the inverse function, for example \(\theta=\tan^{-1}\!\left(\dfrac{\text{opp}}{\text{adj}}\right)\).

What is an angle of elevation?

The angle measured up from the horizontal to a line of sight. In a right-triangle sketch it is the acute angle at the observer; an angle of depression is measured down from the horizontal.

What are the exact ratios of the special angles?

From the special triangles: \(\sin 30^\circ=\dfrac{1}{2}\), \(\cos 30^\circ=\dfrac{\sqrt3}{2}\), \(\tan 45^\circ=1\), \(\tan 60^\circ=\sqrt3\).

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Symmetry Properties Of Trigonometric Functions

Reviewing Trigonometric Ratios

📖 Theory

How to solve a right-triangle problem

Common pitfalls

Frequently asked questions

Theory