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Year 12 Maths Standard 2 (2027) Trigonometry

Right-Angled Trigonometry

20 practice questions 0 video lessons Theory + worked examples

Master right-angled trigonometry for NSW Year 12 Mathematics Standard 2. Using SOH CAH TOA you label the hypotenuse, opposite and adjacent sides relative to an angle, then choose sine, cosine or tangent to find an unknown side or an unknown angle.

You will learn to pick the correct ratio, find a missing side, find a missing angle with the inverse functions, and give angles to the nearest degree or in degrees and minutes β€” a core Standard 2 skill for heights, distances, ramps, ladders and inclines.

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Theory

Right-angled trigonometry uses the ratios of a right-angled triangle's sides β€” SOH CAH TOA β€” to find a missing side or angle. This Year 12 Standard 2 (NSW) guide shows how to label the sides, choose sine, cosine or tangent, find an unknown side or angle, and give angles to the nearest degree or in degrees and minutes.

Right-angled trigonometry connects the angles and side lengths of a right-angled triangle. Relative to a chosen acute angle \(\theta\), the three sides are the hypotenuse \(H\) (opposite the right angle, and the longest side), the opposite \(O\) (facing \(\theta\)) and the adjacent \(A\) (between \(\theta\) and the right angle).

The three ratios are remembered as SOH CAH TOA: \(\sin\theta=\tfrac{O}{H}\), \(\cos\theta=\tfrac{A}{H}\) and \(\tan\theta=\tfrac{O}{A}\). Choose the ratio that uses the two sides you care about, then solve for the missing side or angle.

To find an angle, form the ratio from two known sides and apply the inverse function (\(\sin^{-1}\), \(\cos^{-1}\) or \(\tan^{-1}\)). Angles are given to the nearest degree or in degrees and minutes, where \(1^\circ = 60'\). This is a core Year 12 Standard 2 (NSW) skill for heights, distances, ramps and inclines.

Right-angled triangle with hypotenuse, opposite and adjacent sidesRight-angled triangle showing the hypotenuse opposite the right angle, and the opposite and adjacent sides named relative to the angle theta. hyp opp adj ΞΈ
Name each side relative to \(\theta\): opposite, adjacent and the hypotenuse (opposite the right angle).
Same triangle measured from the other acute angleThe same right-angled triangle with the angle alpha at the top vertex, where the opposite and adjacent sides swap but the hypotenuse is unchanged. hyp adj opp Ξ±
Measure from the other acute angle \(\alpha\) and opposite and adjacent swap β€” but the hypotenuse never changes.

Relative to the angle \(\theta\), the three trigonometric ratios are:

\[\sin\theta=\dfrac{O}{H}\qquad \cos\theta=\dfrac{A}{H}\qquad \tan\theta=\dfrac{O}{A}\]
sinθ=OH,cosθ=AH,tanθ=OA

To find an angle, take the inverse of the ratio:

\[\theta=\sin^{-1}\!\left(\tfrac{O}{H}\right)\quad \theta=\cos^{-1}\!\left(\tfrac{A}{H}\right)\quad \theta=\tan^{-1}\!\left(\tfrac{O}{A}\right)\]
θ=tan1(OA)
Degrees and minutes. \(1^\circ = 60'\). To convert a decimal degree, multiply only the decimal part by \(60\): \(22.62^\circ = 22^\circ + (0.62\times 60)' \approx 22^\circ 37'\).

How to solve a right-angled triangle

  1. Label the sides relative to the angle: hypotenuse (opposite the right angle), opposite (facing the angle) and adjacent (between them).
  2. Choose the ratio using the two sides that matter (SOH CAH TOA): the pair known-and-unknown for a side, or the two known sides for an angle.
  3. Substitute the known values into the ratio equation.
  4. Solve β€” for a side, multiply or divide; for an angle, apply the inverse (\(\sin^{-1}\), \(\cos^{-1}\), \(\tan^{-1}\)) with the calculator in degree mode.
  5. Round as asked β€” to a decimal place, the nearest degree, or convert the decimal degree to minutes (\(\times 60\)).
Example 1 β€” Find a side (sine)
A guy wire \(20\) cm long makes a \(48^\circ\) angle with the ground. Find the vertical height \(x\) it reaches, to \(1\) decimal place.
Solution

The hypotenuse is known and \(x\) is opposite \(48^\circ\), so use sine.

Right triangle, hypotenuse 20 cm, 48 degree angle, opposite side xRight-angled triangle with hypotenuse 20 cm, a 48 degree angle and the opposite side labelled x. 20 cm x 48Β°
\(\sin 48^\circ\)\(=\)\(\dfrac{x}{20}\)
\(x\)\(=\)\(20\sin 48^\circ\)
\(x\)\(=\)\(14.86\ldots\)
\(\therefore\ x\)\(\approx\)\(14.9\text{ cm}\)
x14.9
Example 2 β€” Find a side (tangent)
A ramp rises at \(15^\circ\) over a horizontal run of \(9\) m. Find the vertical rise \(h\), to \(1\) decimal place.
Solution

The adjacent side is known and \(h\) is opposite \(15^\circ\), so use tangent.

Right triangle, adjacent 9 m, 15 degree angle, opposite side hRight-angled triangle with the horizontal run 9 m, a 15 degree incline and the vertical rise labelled h. 9 m h 15Β°
\(\tan 15^\circ\)\(=\)\(\dfrac{h}{9}\)
\(h\)\(=\)\(9\tan 15^\circ\)
\(h\)\(=\)\(2.41\ldots\)
\(\therefore\ h\)\(\approx\)\(2.4\text{ m}\)
h2.4
Example 3 β€” Find an angle (cosine)
A ladder \(4.8\) m long has its base \(1.5\) m from a wall. Find the angle \(\theta\) it makes with the ground, to the nearest degree.
Solution

The hypotenuse and adjacent side are known, so use cosine and its inverse.

Right triangle, hypotenuse 4.8 m, adjacent 1.5 m, angle thetaRight-angled triangle with hypotenuse 4.8 m (a ladder), the base 1.5 m from the wall (adjacent) and the angle theta with the ground. 4.8 m 1.5 m ΞΈ
\(\cos\theta\)\(=\)\(\dfrac{1.5}{4.8}\)
\(\theta\)\(=\)\(\cos^{-1}\!\left(\dfrac{1.5}{4.8}\right)\)
\(\theta\)\(=\)\(71.79\ldots^\circ\)
\(\therefore\ \theta\)\(\approx\)\(72^\circ\)
θ72°
Example 4 β€” Angle in degrees and minutes
A path rises \(5\) m for every \(12\) m measured horizontally. Find the angle \(\theta\) with the horizontal, to the nearest minute.
Solution

Use tangent, then convert the decimal degree to minutes.

Right triangle, rise 5 m, run 12 m, angle thetaRight-angled triangle with a vertical rise of 5 m, a horizontal run of 12 m and the angle theta the path makes with the horizontal. 5 m 12 m ΞΈ
\(\tan\theta\)\(=\)\(\dfrac{5}{12}\)
\(\theta\)\(=\)\(\tan^{-1}\!\left(\dfrac{5}{12}\right)\)
\(\theta\)\(=\)\(22.62\ldots^\circ\)
\(0.62\ldots^\circ\)\(=\)\(0.62\ldots\times 60'\approx 37'\)
\(\therefore\ \theta\)\(\approx\)\(22^\circ 37'\)
θ22°37

Common pitfalls

Re-label when the angle changes. Opposite and adjacent are named relative to the angle you are using; if you switch to the other acute angle they swap. Only the hypotenuse (opposite the right angle) stays fixed.
Use the inverse for angles. To find an angle you need \(\sin^{-1}\), \(\cos^{-1}\) or \(\tan^{-1}\) β€” not \(\sin\), \(\cos\), \(\tan\) β€” and the calculator must be in degree mode.
Convert minutes correctly. Multiply only the decimal part of the degree by \(60\): \(22.62^\circ = 22^\circ 37'\), never \(22^\circ 62'\). Minutes never reach \(60\).

Frequently asked questions

What does SOH CAH TOA mean?

It is a memory aid for the three trigonometric ratios in a right-angled triangle: Sine equals Opposite over Hypotenuse, Cosine equals Adjacent over Hypotenuse, and Tangent equals Opposite over Adjacent. You pick the one that uses the two sides you know or want.

How do I know whether to use sin, cos or tan?

Label the three sides relative to the angle, then look at which two sides are involved. Opposite and hypotenuse means sine, adjacent and hypotenuse means cosine, opposite and adjacent means tangent. That is exactly SOH CAH TOA.

How do I find an unknown angle?

Form the ratio from the two known sides, for example tan of the angle equals opposite over adjacent, then apply the inverse function on your calculator (tan to the minus one). Make sure the calculator is in degree mode, then round to the nearest degree or convert to minutes.

How do I convert a decimal degree into degrees and minutes?

Keep the whole-number degrees, then multiply only the decimal part by 60 to get the minutes. For example 22.62 degrees becomes 22 degrees and 0.62 times 60, which is about 37 minutes, so 22 degrees 37 minutes. Minutes are always less than 60.

Which side is the hypotenuse?

The hypotenuse is always the side opposite the right angle, and it is the longest side of a right-angled triangle. It stays the hypotenuse no matter which acute angle you are working with; only the opposite and adjacent sides change.