Mathematics Standard • Year 11 • Module 2 • Lesson 15

Right-Angled Trig: Finding Angles — Skill Drill

Build fluency with inverse trig: pick the right ratio from two known sides, apply sin⁻¹/cos⁻¹/tan⁻¹, and convert to degrees and minutes.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Match each pair of known sides to the inverse trig function.

O and H known → θ = ____________ A and H known → θ = ____________ O and A known → θ = ____________

Q1.2 The two acute angles of a right-angled triangle add to ____________ ° (they are complementary).

Q1.3 Convert 30.26° to degrees and minutes (round to the nearest minute): ____________ ° ____________ '

Stuck? Revisit lesson § Inverse Trigonometric Functions and § Degrees and Minutes conversion.

2. Worked example — inverse sine, then convert to degrees/minutes

Follow each line of working. Every step has a reason on the right.

Problem. A right-angled triangle has O = 7 cm and H = 11 cm. Find the angle θ opposite the 7 cm side, in degrees and minutes.

Step 1 — Identify ratio from known sides.

sin θ = O ÷ H = 7 ÷ 11 = 0.6364...

Reason: O and H are known → sine.

Step 2 — Apply inverse sine.

θ = sin⁻¹(0.6364) ≈ 39.52°

Reason: inverse sine gives the angle whose sine is 0.6364. Calculator: [SHIFT] [sin].

Step 3 — Convert decimal to minutes.

0.52 × 60 = 31.2 ≈ 31'; θ ≈ 39°31'

Reason: multiply the decimal part by 60 to get minutes; round to the nearest whole minute.

Conclusion. θ ≈ 39°31'.

3. Faded example — inverse tangent, then minutes

A building is 45 m tall. An observer stands 60 m from the base. Find the angle of elevation to the top, in degrees and minutes. Fill in each blank. 4 marks

Step 1 — Identify ratio:

O = ____________ m, A = ____________ m. Use tan θ = O ÷ A = ____________ ÷ ____________ = ____________

Step 2 — Apply inverse tangent:

θ = tan⁻¹(____________) ≈ ____________ °

Step 3 — Convert decimal to minutes:

____________ × 60 ≈ ____________ ' ⇒ θ ≈ ____________ ° ____________ '

Conclusion. The angle of elevation is ____________ ° ____________ '.

Stuck? Revisit lesson § Worked Example 2 — Angle of Elevation.

4. Graduated practice — Finding angles

Express all answers in degrees and minutes unless told otherwise. Show the ratio, the inverse, and the conversion.

Foundation — single inverse (4 questions)

Q	Problem	Answer
4.1 1	O = 5, H = 13. Find θ.
4.2 1	A = 9, H = 15. Find θ.
4.3 1	O = 7, A = 24. Find θ.
4.4 1	O = 4.2 cm, H = 9.8 cm. Find θ.

Standard — typical HSC difficulty (6 questions)

Show your ratio, the calculator step and the conversion to minutes.

4.5 A = 8.5 m, H = 14.0 m. Find θ. 2 marks

4.6 A right-angled triangle has legs 5 m and 12 m. Find both acute angles. 2 marks

4.7 A right-angled triangle has hypotenuse 25 cm and one leg 7 cm. Find both acute angles. 2 marks

4.8 A ramp rises 1.5 m over a horizontal distance of 9 m. Find the angle the ramp makes with the horizontal. 2 marks

4.9 From a lighthouse 48 m above sea level, a boat is 120 m horizontally from the base. Find the angle of depression to the boat. 2 marks

4.10 A ski slope descends 280 m vertically over a horizontal distance of 650 m. Find the angle of inclination. 2 marks

Extension — combine with side calculation (2 questions)

4.11 A wire is attached from the top of a 12 m pole to the ground, 5 m from the base. Find the angle the wire makes with the ground. 3 marks

4.12 A ship travels 40 km north then 30 km east. Find the angle its overall path deviates from due north (i.e. the bearing east of north), in degrees and minutes. 3 marks

Stuck on 4.11? With pole = 12 m (O) and ground distance = 5 m (A), use tan⁻¹(12/5).

5. Self-check the easy 3

Tick the first three once you've checked your method works.

For 4.1 I used sin⁻¹ (O and H known), not tan⁻¹ or cos⁻¹.

For 4.2 I used cos⁻¹ (A and H known).

For 4.3 I used tan⁻¹ (O and A known, no H involved).

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Inverse trig matches

O and H → θ = sin⁻¹(O/H). A and H → θ = cos⁻¹(A/H). O and A → θ = tan⁻¹(O/A).

Q1.2 — Sum of acute angles

The two acute angles sum to 90°.

Q1.3 — 30.26° to ° '

0.26 × 60 = 15.6 → 16'. So 30.26° ≈ 30°16'.

Q3 — Faded example (building 45 m, observer 60 m)

Step 1: O = 45, A = 60. tan θ = 45 ÷ 60 = 0.75.
Step 2: θ = tan⁻¹(0.75) ≈ 36.87°.
Step 3: 0.87 × 60 ≈ 52' → θ ≈ 36°52'.
Conclusion: angle of elevation ≈ 36°52'.

Q4.1 — sin⁻¹(5/13)

sin⁻¹(0.3846) ≈ 22.62° = 22°37'.

Q4.2 — cos⁻¹(9/15)

cos⁻¹(0.6) ≈ 53.13° = 53°8'.

Q4.3 — tan⁻¹(7/24)

tan⁻¹(0.2917) ≈ 16.26° = 16°16'.

Q4.4 — sin⁻¹(4.2/9.8)

sin⁻¹(0.4286) ≈ 25.38° = 25°23'.

Q4.5 — cos⁻¹(8.5/14)

cos⁻¹(0.6071) ≈ 52.61° = 52°37'.

Q4.6 — Legs 5 and 12

α = tan⁻¹(5/12) ≈ 22.62° = 22°37'. β = 90° − 22°37' = 67°23'.

Q4.7 — H = 25, leg = 7

α = sin⁻¹(7/25) ≈ 16.26° = 16°16'. β = 90° − 16°16' = 73°44'.

Q4.8 — Ramp rises 1.5 m over 9 m

tan⁻¹(1.5/9) = tan⁻¹(0.1667) ≈ 9.46° = 9°28'.

Q4.9 — Lighthouse 48 m, boat 120 m horizontal

Angle of depression = tan⁻¹(48/120) = tan⁻¹(0.4) ≈ 21.80° = 21°48'.

Q4.10 — Ski slope

tan⁻¹(280/650) ≈ tan⁻¹(0.4308) ≈ 23.32° = 23°19'.

Q4.11 — Pole 12 m, anchor 5 m from base

tan⁻¹(12/5) = tan⁻¹(2.4) ≈ 67.38° = 67°23' (angle the wire makes with the ground).

Q4.12 — Ship 40 N then 30 E

Path deviates east of north by tan⁻¹(30/40) = tan⁻¹(0.75) ≈ 36.87° = 36°52' east of north.