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.