Mathematics Standard • Year 11 • Module 2 • Lesson 14

Right-Angled Trig: Finding Sides — Skill Drill

Build fluency with SOHCAHTOA for finding unknown sides — labelling O/A/H, choosing the right ratio, and handling both numerator and denominator unknowns.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete each SOHCAHTOA ratio.

sin θ = ____ ÷ ____ cos θ = ____ ÷ ____ tan θ = ____ ÷ ____

Q1.2 If the unknown side is in the numerator (e.g. sin θ = x/H), the formula to solve for x is: x = ____________ .

If the unknown side is in the denominator (e.g. sin θ = O/x), the formula is: x = ____________ .

Q1.3 Convert 35°30' to decimal degrees: ____________ ° Convert 47°12' to decimal degrees: ____________ °

Stuck? Revisit lesson § Always Label First and § Worked Example 4 — Degrees and Minutes.

2. Worked example — unknown in numerator (use sin)

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

Problem. A right-angled triangle has hypotenuse 18 m and one angle 38°. Find the side opposite the 38° angle, correct to 2 d.p.

Step 1 — Label sides and choose ratio.

Known: H = 18 m, θ = 38°. Need: O. Use sin θ = O ÷ H.

Reason: O and H are involved → sine (S-O-H).

Step 2 — Rearrange (unknown in numerator).

O = H × sin θ = 18 × sin 38°

Reason: multiply both sides by H; unknown is in the numerator.

Step 3 — Evaluate and round.

O = 18 × 0.6157... ≈ 11.08 m

Reason: calculator gives sin 38° ≈ 0.6157; round to 2 d.p.

Conclusion. The opposite side is 11.08 m.

3. Faded example — unknown in denominator (use cos)

A right-angled triangle has angle 52° and the adjacent side is 9.4 cm. Find the hypotenuse, correct to 2 d.p. Fill in each blank. 4 marks

Step 1 — Label and choose ratio:

Known: A = ____________ , θ = ____________ . Need: H. Use cos θ = ____ ÷ ____ .

Step 2 — Substitute and rearrange:

cos 52° = ____________ ÷ H ⇒ H = ____________ ÷ cos 52°

Step 3 — Evaluate:

H = 9.4 ÷ ____________ ≈ ____________ cm (to 2 d.p.)

Conclusion. The hypotenuse is ____________ cm.

Stuck? Revisit lesson § Worked Example 2 — Unknown in Denominator.

4. Graduated practice — Finding sides

For every question: state the ratio chosen, show substitution, evaluate, then round as told.

Foundation — single-step ratios (4 questions)

Q	Problem	Answer (to 2 d.p.)
4.1 1	θ = 42°, H = 15 cm. Find O.
4.2 1	θ = 28°, H = 22 m. Find A.
4.3 1	θ = 55°, A = 8.4 cm. Find O.
4.4 1	θ = 36°, O = 7 m. Find H.

Standard — typical HSC difficulty (6 questions)

Show the ratio, a line of substitution and a clearly labelled final answer with units. All to 2 d.p. unless told otherwise.

4.5 θ = 71°, A = 5.5 m. Find H. 2 marks

4.6 θ = 48°, O = 12 cm. Find A. 2 marks

4.7 θ = 35°30', H = 20 m. Find O. 2 marks

4.8 A ramp makes an angle of 15° with the horizontal. The ramp is 8 m long along its slope. Find its vertical rise. 2 marks

4.9 A flagpole casts a 12 m shadow when the sun's angle of elevation is 54°. Find the flagpole's height to 1 d.p. 2 marks

4.10 A cable car wire goes from ground level to a mountain station 420 m higher. The wire makes an angle of 28° with the horizontal. Find the length of the wire to the nearest metre. 2 marks

Extension — bearings and depression (2 questions)

4.11 From the top of a 25 m cliff, the angle of depression to a boat is 18°. Find the horizontal distance from the base of the cliff to the boat, to 1 d.p. 3 marks

4.12 A ladder makes an angle of 63°24' with the ground and reaches 4.6 m up a vertical wall. Find the length of the ladder, to 2 d.p. 3 marks

Stuck on 4.11? Angle of depression at the top = angle of elevation at the boat (alternate angles). Redraw with the 18° at the boat: O = 25 m, A = horizontal distance, use tan.

5. Self-check the easy 3

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

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

For 4.2 I labelled A as the side next to the 28° angle (and not the hypotenuse), and used cos.

For 4.4 I noticed the unknown H is in the denominator of sin θ = O/H, so I divided (H = O ÷ sin θ) instead of multiplying.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — SOHCAHTOA

sin θ = O ÷ H cos θ = A ÷ H tan θ = O ÷ A.

Q1.2 — Rearranging

Numerator unknown: x = H × sin θ (or appropriate side × the chosen ratio). Denominator unknown: x = O ÷ sin θ (or the known side ÷ the ratio).

Q1.3 — Degrees and minutes

35°30' = 35 + 30/60 = 35.5°. 47°12' = 47 + 12/60 = 47.2°.

Q3 — Faded example (θ = 52°, A = 9.4)

Step 1: Known A = 9.4 cm, θ = 52°. Ratio: cos θ = A ÷ H.
Step 2: cos 52° = 9.4 ÷ H ⇒ H = 9.4 ÷ cos 52°.
Step 3: H = 9.4 ÷ 0.6157 ≈ 15.27 cm.
Conclusion: H ≈ 15.27 cm.

Q4.1 — θ=42°, H=15

O = 15 × sin 42° ≈ 15 × 0.6691 ≈ 10.04 cm.

Q4.2 — θ=28°, H=22

A = 22 × cos 28° ≈ 22 × 0.8829 ≈ 19.43 m.

Q4.3 — θ=55°, A=8.4

tan 55° = O ÷ 8.4 ⇒ O = 8.4 × tan 55° ≈ 8.4 × 1.4281 ≈ 12.00 cm.

Q4.4 — θ=36°, O=7

sin 36° = 7 ÷ H ⇒ H = 7 ÷ sin 36° ≈ 7 ÷ 0.5878 ≈ 11.90 m.

Q4.5 — θ=71°, A=5.5

cos 71° = 5.5 ÷ H ⇒ H = 5.5 ÷ cos 71° ≈ 5.5 ÷ 0.3256 ≈ 16.89 m.

Q4.6 — θ=48°, O=12

tan 48° = 12 ÷ A ⇒ A = 12 ÷ tan 48° ≈ 12 ÷ 1.1106 ≈ 10.80 cm.

Q4.7 — θ=35°30', H=20

35°30' = 35.5°. O = 20 × sin 35.5° ≈ 20 × 0.5807 ≈ 11.61 m.

Q4.8 — Ramp at 15°, slope 8 m, vertical rise

Slope is the hypotenuse; rise = O. sin 15° = O ÷ 8 ⇒ O = 8 × sin 15° ≈ 8 × 0.2588 ≈ 2.07 m.

Q4.9 — Flagpole with 12 m shadow, 54° elevation

tan 54° = h ÷ 12 ⇒ h = 12 × tan 54° ≈ 12 × 1.3764 ≈ 16.5 m.

Q4.10 — Cable car wire

O = 420 m, θ = 28°, find H. sin 28° = 420 ÷ H ⇒ H = 420 ÷ sin 28° ≈ 420 ÷ 0.4695 ≈ 894 m.

Q4.11 — Cliff 25 m, depression 18°

Alternate angles: 18° also the elevation at the boat. With O = 25 m, θ = 18°, A = d: tan 18° = 25 ÷ d ⇒ d = 25 ÷ tan 18° ≈ 25 ÷ 0.3249 ≈ 76.9 m.

Q4.12 — Ladder at 63°24', wall reach 4.6 m

63°24' = 63 + 24/60 = 63.4°. Wall reach is O; ladder length is H. sin 63.4° = 4.6 ÷ H ⇒ H = 4.6 ÷ sin 63.4° ≈ 4.6 ÷ 0.8942 ≈ 5.15 m.