Mathematics • Year 10 • Unit 3 • Lesson 2

Finding Unknown Sides in the Real World

Apply Lesson 2 to real Australian situations — a leaning ladder against a fence in Hurstville, the supports of a Western Plains solar panel, a wheelchair ramp at a Newcastle library. Each problem asks for an unknown length using sin, cos or tan. Round to 2 decimal places unless told otherwise. Calculator in DEG mode.

Apply · Real-World Maths

1. Word problems

For each problem: (i) sketch the right-angled triangle, (ii) identify the marked angle, given side and unknown side, (iii) write the ratio, (iv) rearrange and solve. A correct final answer with no working only earns half marks.

1.1 — Hurstville ladder. A 4.5 m ladder leans against a vertical wall. It makes an angle of 72° with the ground.

(a) How high up the wall does the ladder reach? (Round to 2 d.p.)
(b) How far is the foot of the ladder from the wall? 4 marks

Stuck? Ladder = hypotenuse = 4.5. Wall height = opposite (use sin). Distance from wall = adjacent (use cos).

1.2 — Solar panel tilt. A solar panel on a Western Plains farm is mounted on a frame so the panel makes an angle of 35° with the horizontal ground. The vertical leg of the frame (rising from the ground to the back edge of the panel) is 1.2 m.

How long is the sloping panel? Round to 2 d.p. 3 marks

Stuck? The panel is the hypotenuse. The vertical leg (1.2 m) is opposite the 35° angle. sin 35° = 1.2 / panel length.

1.3 — Wheelchair ramp. The maximum legal slope for an Australian wheelchair ramp is 1 : 14 (a ramp angle of about 4.09°). A new ramp at a Newcastle library has a vertical rise of 0.75 m and is built at the maximum legal angle.

(a) Find the horizontal length the ramp covers (the adjacent side). Round to 2 d.p.
(b) Find the length of the ramp itself (the hypotenuse). 4 marks

Stuck? Vertical rise = opposite = 0.75 m. Angle = 4.09°. For horizontal use tangent. For ramp length use sine.

1.4 — Tent guy rope. A guy rope from the top of a 1.8 m tent pole is pegged into the ground 1.1 m away from the base of the pole. (Assume the pole is vertical and the rope is straight and taut.)

(a) Use tan to find the angle the rope makes with the ground. (Round to the nearest whole degree — yes, this previews Lesson 3.)
(b) Use sin (with the angle from part (a)) to find the length of the rope. Round to 2 d.p. 3 marks

Stuck? tan θ = 1.8 / 1.1, so θ = inverse tan of (1.8 / 1.1). Then rope = 1.8 / sin θ.

1.5 — Roof rafter. The pitched roof of a Queenslander cottage has a triangular cross-section. One rafter sits at 30° to the horizontal ceiling. The horizontal distance from the wall to directly under the ridge is 4.2 m.

(a) Use the exact-value table to find the height of the ridge above the ceiling. Leave the answer in exact form (no calculator).
(b) Use the exact-value table to find the length of the rafter. Leave in exact form. 3 marks

Stuck? Adjacent = 4.2 m, angle = 30°. tan 30° = 1/√3 (height / 4.2). cos 30° = √3 / 2 (4.2 / rafter).

2. Explain your thinking

This question is about communication, not just numbers. Use full sentences. 4 marks

2.1 A classmate sets up the following problem and asks for your help: "Hypotenuse 12, angle 30°, find the side adjacent to the 30°. I wrote sin 30° = x / 12, so x = 12 × sin 30° = 6." Explain in 4-6 sentences (i) what they did right, (ii) the specific error they made (referring to SOH-CAH-TOA explicitly), (iii) what the correct ratio should have been, and (iv) the corrected final answer. Include the words "match the pair".

Stuck? They used sine, which pairs opposite + hypotenuse — but the question asked for the adjacent side. They needed cosine.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — Hurstville ladder

(a) sin 72° = wall / 4.5 → wall = 4.5 × sin 72° = 4.5 × 0.9511 = 4.2799 ≈ 4.28 m.
(b) cos 72° = base / 4.5 → base = 4.5 × cos 72° = 4.5 × 0.3090 = 1.3906 ≈ 1.39 m.
Pythagoras sanity: 4.28² + 1.39² ≈ 18.32 + 1.93 = 20.25 = 4.5². ✓

1.2 — Solar panel

sin 35° = 1.2 / panel → panel = 1.2 / sin 35° = 1.2 / 0.5736 = 2.0921 ≈ 2.09 m.
The panel is the hypotenuse — and (as expected) it is longer than the 1.2 m vertical leg.

1.3 — Wheelchair ramp

(a) tan 4.09° = 0.75 / horizontal → horizontal = 0.75 / tan 4.09° = 0.75 / 0.07153 = 10.4848 ≈ 10.48 m.
(b) sin 4.09° = 0.75 / ramp → ramp = 0.75 / sin 4.09° = 0.75 / 0.07133 = 10.5147 ≈ 10.51 m.
The ramp is only marginally longer than the horizontal because the angle is so shallow.

1.4 — Tent guy rope

(a) tan θ = 1.8 / 1.1 = 1.6364, so θ = inverse tan (1.6364) ≈ 58.57° ≈ 59°.
(b) sin 58.57° = 1.8 / rope → rope = 1.8 / sin 58.57° = 1.8 / 0.8534 = 2.1093 ≈ 2.11 m.
Pythagoras check: 1.8² + 1.1² = 3.24 + 1.21 = 4.45, √4.45 ≈ 2.11. ✓

1.5 — Roof rafter (exact values)

(a) tan 30° = height / 4.2, so height = 4.2 × tan 30° = 4.2 × (1/√3) = 4.2 / √3 = 4.2√3 / 3 m (≈ 2.42 m).
(b) cos 30° = 4.2 / rafter, so rafter = 4.2 / cos 30° = 4.2 / (√3 / 2) = 8.4 / √3 = 8.4√3 / 3 m (≈ 4.85 m).
Exact-value answers come straight from the lesson's table — no calculator needed.

2.1 — Explain your thinking (sample response)

My classmate set up their angle and given side correctly — they spotted the 30° and the 12, and they made the unknown the subject. The error is that they used sine, but SOH says sine pairs opposite and hypotenuse. The question asked for the side adjacent to the 30° angle, so we needed the pair "adjacent + hypotenuse", which is the CAH ratio — cosine. The Lesson 2 rule is "match the pair" — list the two sides involved in the problem, then read off the ratio that connects them. Corrected working: cos 30° = x / 12, so x = 12 × cos 30° = 12 × (√3 / 2) ≈ 10.39. (My classmate's answer of 6 was for the opposite side, not the adjacent.)

Marking: 1 for praising the angle/given identification, 1 for naming the SOH-vs-CAH confusion, 1 for the corrected formula, 1 for the corrected numerical answer.