Right-Angle Triangle Trigonometry

Key Terms

SOH-CAH-TOA: sin θ = Opp/Hyp; cos θ = Adj/Hyp; tan θ = Opp/Adj.
Hypotenuse: The longest side of a right-angled triangle; always opposite the right angle.
Inverse trig functions: sin²θ, cos²θ, tan²θ give the angle from a known ratio: e.g. θ = sin²(0.5) = 30°.
Angle of elevation: The angle measured upward from the horizontal to a line of sight to an object above.
Angle of depression: The angle measured downward from the horizontal to a line of sight to an object below; equals the angle of elevation from the object.
Exact values: sin 30° = ½, cos 30° = √3/2, tan 30° = 1/√3; sin 45° = 1/√2; sin 60° = √3/2, tan 60° = √3.

Labelling a Right-Angle Triangle

Before applying any trigonometric ratio, you must correctly label the three sides relative to the angle of interest θ.

Key Rule: Always identify the Hypotenuse first — it is the longest side, directly opposite the right angle. Then label Opposite (O) as the side directly across from θ, and Adjacent (A) as the side next to θ (that is not H).

SOH-CAH-TOA

SOH

\(\sin\theta = \dfrac{O}{H}\)

CAH

\(\cos\theta = \dfrac{A}{H}\)

TOA

\(\tan\theta = \dfrac{O}{A}\)

QCAA Formula Sheet: sin θ = O/H | cos θ = A/H | tan θ = O/A

Finding a Missing Side

Identify which ratio connects the known angle, the known side, and the unknown side. Then substitute and solve.

Known angle and H, want O → use sin: O = H sin θ
Known angle and A, want O → use tan: O = A tan θ
Known angle and O, want H → use sin: H = O / sin θ

Finding a Missing Angle

Use the inverse trig functions on your calculator:

\[\theta = \sin^{-1}\!\left(\frac{O}{H}\right) \qquad \theta = \cos^{-1}\!\left(\frac{A}{H}\right) \qquad \theta = \tan^{-1}\!\left(\frac{O}{A}\right)\]

Calculator Check: Always ensure your calculator is in DEGREES mode before computing any trig ratio. The answer will be wrong if it is in radians.

Worked Example 1 — Finding a Side

Problem: A right-angled triangle has an angle of 35° and a hypotenuse of 14 cm. Find the side opposite the 35° angle.

Step 1: Label sides. We have θ = 35°, H = 14 cm, want O.

Step 2: Identify ratio. O and H → use sin.

Step 3: Write equation and solve.

\[\sin 35° = \frac{O}{14}\] \[O = 14 \times \sin 35° = 14 \times 0.5736\ldots \approx \mathbf{8.03 \text{ cm}}\]

Answer: The opposite side is approximately 8.03 cm.

Worked Example 2 — Finding an Angle

Problem: A right-angled triangle has legs of 5 cm and 8 cm. Find the acute angle between the hypotenuse and the base (8 cm leg).

Step 1: Label sides relative to the required angle θ. The base 8 cm is adjacent, the other leg 5 cm is opposite.

Step 2: O and A are known → use tan.

\[\tan\theta = \frac{O}{A} = \frac{5}{8}\] \[\theta = \tan^{-1}\!\left(\frac{5}{8}\right) = \tan^{-1}(0.625) \approx \mathbf{32.0°}\]

Answer: The angle is approximately 32.0°.

Hot Tip: Before writing any equation, always identify H first (it is opposite the right angle and is the longest side). Then decide which of the remaining two sides is O and which is A based on the angle you know or want. Once labelled correctly, SOH-CAH-TOA gives you the right formula immediately.

Why Does Trigonometry Work?

All right-angled triangles with the same angles are similar — their sides are in fixed ratios regardless of size. Trigonometry names these fixed ratios. For any angle θ in a right triangle, the ratio O/H is always the same number, no matter how large or small the triangle. We call this fixed ratio sin θ.

This is why a calculator can tell you sin 35° = 0.5736… — that single number describes the ratio O/H for every right triangle with a 35° angle.

Step-by-Step Method

Draw and label the triangle. Mark the right angle. Identify the angle θ you are using.
Label sides: H (opposite right angle), O (opposite θ), A (adjacent to θ, not H).
Identify which two sides are involved (known + unknown) and choose the correct ratio.
Write the equation, then solve algebraically for the unknown.
Check: is your answer reasonable? (Hypotenuse must be the longest side.)

Memory Tricks

Some people remember SOH-CAH-TOA as: "Some Old Hens Cackle And Howl Throughout Our Area" or simply say the word “soh-cah-toa” aloud.

Guide Example

Problem: Find the hypotenuse of a right triangle if one angle is 52° and the adjacent side is 9 m.

We have θ = 52°, A = 9 m, want H.

A and H are involved → use cos:

\[\cos 52° = \frac{A}{H} = \frac{9}{H}\] \[H = \frac{9}{\cos 52°} = \frac{9}{0.6157\ldots} \approx 14.62 \text{ m}\]

Sanity check: H = 14.62 m > A = 9 m. ✓

Common Mistake: Students often write sin θ = O/H, then solve incorrectly as O = H/sin θ (dividing instead of multiplying). Use basic algebra: if sin θ = O/H, then O = H × sin θ. If you want H: H = O / sin θ. The ratio stays the same — just rearrange carefully.

Pythagoras vs Trigonometry

Use Pythagoras (a² + b² = c²) when you know two sides and want the third. Use trigonometry when you know an angle and a side, or when you want an angle.

Mastery Practice

See Answers ➔

Fluency Find the marked side, correct to 2 decimal places.
1. A right triangle has angle θ = 42° and hypotenuse 18 cm. Find the opposite side.
2. A right triangle has angle θ = 67° and adjacent side 11 m. Find the hypotenuse.
3. A right triangle has angle θ = 28° and hypotenuse 25 cm. Find the adjacent side.
View Solution
Fluency Find the marked angle to the nearest degree.
1. Opposite = 7, hypotenuse = 15.
2. Adjacent = 8, hypotenuse = 13.
3. Opposite = 12, adjacent = 9.
View Solution
Fluency A right-angled triangle has legs of 6 cm and 10 cm. Find:
1. The length of the hypotenuse.
2. Both acute angles (to the nearest degree).
View Solution
Understanding A ladder 8 m long leans against a wall. The base of the ladder is 2.5 m from the wall. Find:
1. The angle the ladder makes with the ground.
2. The height the ladder reaches up the wall.
View Solution
Understanding A ramp rises 1.2 m over a horizontal distance of 6 m. Find:
1. The angle of inclination of the ramp.
2. The length of the ramp surface.
View Solution
Understanding From a point 50 m from the base of a building, the angle of elevation to the top of the building is 38°. Find the height of the building, correct to 2 decimal places. View Solution
Understanding A guy wire is attached from the top of a 12 m vertical pole to a point on the ground 9 m from the base of the pole. Find:
1. The length of the wire.
2. The angle the wire makes with the ground.
View Solution
Understanding A park path runs diagonally across a rectangular park that is 80 m long and 60 m wide. Find:
1. The length of the diagonal path.
2. The angle the path makes with the 80 m side.
View Solution
Problem Solving A triangle ABC has a right angle at C, with AC = 7 cm and BC = 24 cm. Find:
1. The length of AB (the hypotenuse).
2. Angle ABC (to the nearest degree).
3. Angle BAC (to the nearest degree).
4. Verify that the three angles sum to 180°.
View Solution
Problem Solving Two vertical poles are 20 m apart. Pole A is 8 m high; Pole B is 15 m high. A wire runs from the top of Pole A to the base of Pole B, and another wire runs from the top of Pole B to the base of Pole A. The two wires cross at point X.
1. Find the angle that each wire makes with the ground.
2. Let d be the horizontal distance from Pole A to the crossing point X. Using the two wire slopes, set up an equation in d and solve it.
3. Find the height of the crossing point X above the ground.
Hint: express the height of X as a function of horizontal position using tan for each wire, then equate the two expressions.
View Solution