L3 · Finding an angle: inverse ratios

Learning goal

“I can find an unknown acute angle of a right triangle when I know two sides: I build the ratio from the two known sides, then use the inverse key (sin⁻¹, cos⁻¹, tan⁻¹) to undo the ratio and recover the angle.”

Inverse trigonometric ratio — the calculator operation that undoes a ratio: sin⁻¹, cos⁻¹, tan⁻¹. Feed it a ratio, get back the angle.

Ratio value — the decimal you get by dividing the two sides before touching the inverse key, e.g. 6 ÷ 12 = 0.5.

Acute angle — an angle between 0° and 90°; both non-right angles of a right triangle are acute, so every angle answer in this unit lands in that range.

Worked example

A right triangle has hypotenuse 12 cm. The side opposite the marked angle θ is 6 cm. Find θ. Step 1: Sides in play: opposite and hypotenuse -> sin. Step 2: Build the ratio value first, as an ordinary division: sin θ = 6/12 = 0.5. Step 3: Undo it. θ = sin⁻¹(0.5) = 30°. Say it plainly: the calculator's sin⁻¹ does not "do trigonometry" — it answers one question: "sin of what angle equals this number?"

Check yourself — explain it

Why must the division 6 ÷ 12 happen before the inverse key?

Paired problem — example still visible

Hypotenuse 10 cm, side adjacent to θ is 8 cm. Find θ -> 37° (cos θ = 8/10 = 0.8, so θ = cos⁻¹(0.8) = 37°)

Faded completion — last step is yours

"Opposite 5 cm, adjacent 12 cm: tan θ = 5/12 ≈ 0.417, so θ = tan⁻¹(0.417) = ___°" -> 23°

Ready to practise?

10 scored questions with a 3-rung hint ladder. Mastery needs 8/10 plus both gate questions.

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Grade 10 — Mathematics 10C › C10U03 Trigonometry of Right Triangles

L3 · Finding an angle: inverse ratios