Mathematics • Year 10 • Unit 3 • Lesson 1
Trig Ratios in Real Triangles
Apply SOH CAH TOA to real Australian contexts — a ladder against a Sydney terrace wall, a kite string at Bondi, the slope of a Snowy-Mountains ski run. In Lesson 1 you only write the ratios; finding numerical sides comes in Lesson 2. So every problem here asks "which ratio?" before any number-crunching.
1. Word problems
Each problem describes a real right-angled triangle. For each one: (i) sketch the triangle and label H, O, A relative to the marked angle θ, (ii) write the relevant trig ratio for θ as a fraction, and (iii) simplify if possible. Do not calculate side lengths or angles — that is Lessons 2 and 3.
1.1 — Ladder against a wall. A 5 m ladder is propped against a vertical wall in Newtown. The foot of the ladder is 3 m from the wall and the top of the ladder reaches 4 m up the wall. Let θ be the angle the ladder makes with the ground.
(a) Sketch the triangle and label H, O, A relative to θ.
(b) Write sin θ, cos θ and tan θ as fractions in simplest form. 4 marks
1.2 — Kite at Bondi. A kite-flyer at Bondi has let out 25 m of string. The kite is directly above a point 7 m along the sand from where she is standing, and 24 m above the sand. The string is straight. Let θ be the angle the string makes with the horizontal sand.
(a) Identify H, O and A relative to θ.
(b) Write sin θ as a fraction in simplest form. 3 marks
1.3 — Ski run gradient. A short straight ski run at Thredbo drops 60 m vertically over a horizontal distance of 80 m. The straight-line slope length is 100 m. Let θ be the angle the slope makes with the horizontal.
(a) Write tan θ as a fraction in simplest form. (Use the side ratio that does not involve the hypotenuse.)
(b) Write sin θ and cos θ as fractions in simplest form. 3 marks
1.4 — Sydney Harbour Bridge cable. A diagonal support cable runs from a point 9 m above the deck down to a point on the deck 12 m horizontally away from the foot of the support. The cable itself is 15 m long. Let θ be the angle the cable makes with the deck.
(a) Identify H, O, A relative to θ.
(b) Write cos θ as a fraction in simplest form. 3 marks
1.5 — Surveyor's right-angled set-square. A surveyor in Parramatta uses a 30-60-90 set-square (sides 1 unit, √3 units, 2 units — the lesson's "second special triangle"). The 60° angle is at one corner.
(a) Without a calculator, write down sin 60°, cos 60° and tan 60° as exact values.
(b) Confirm that sin 60° = cos 30° using only the exact-value table from the lesson. 3 marks
2. Explain your thinking
This question is about communication, not just numbers. Use full sentences. 4 marks
2.1 A friend tells you: "I just memorise that the side at the bottom of a triangle is always the adjacent." Using the labels and rules from Lesson 1, explain in 4-6 sentences (i) why that rule is wrong, (ii) what actually determines which side is the adjacent, and (iii) what happens to the O and A labels if the same triangle is rotated 90°. Refer to the words "marked angle" and "hypotenuse" somewhere in your answer.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Ladder (3-4-5)
(a) H = 5 m (ladder), O = 4 m (wall, across from θ at the ground), A = 3 m (along the ground, next to θ).
(b) sin θ = 4/5, cos θ = 3/5, tan θ = 4/3. Classic 3-4-5 triangle.
1.2 — Kite string (7-24-25)
(a) H = 25 m (string), O = 24 m (height of the kite, across from θ at the sand), A = 7 m (horizontal distance, next to θ).
(b) sin θ = O / H = 24/25. Already in simplest form — gcd(24, 25) = 1.
1.3 — Ski slope (60-80-100)
This is a 3-4-5 triangle ×20.
(a) tan θ = O / A = 60 / 80 = 3/4.
(b) sin θ = 60 / 100 = 3/5. cos θ = 80 / 100 = 4/5.
"Gradient" in everyday language is the tan of the slope angle.
1.4 — Bridge cable (9-12-15)
(a) H = 15 m (cable), O = 9 m (vertical rise, across from θ), A = 12 m (horizontal, next to θ at the deck).
(b) cos θ = A / H = 12 / 15 = 4/5. Another scaled 3-4-5 triangle (×3).
1.5 — Surveyor's 30-60-90 set-square
(a) From the lesson's exact-value table:
sin 60° = √3 / 2, cos 60° = 1/2, tan 60° = √3.
(b) The table also gives cos 30° = √3 / 2, which equals sin 60°. ✓
This co-function pattern — sin (x°) = cos (90° − x°) — comes back constantly in senior Maths.
2.1 — Explain your thinking (sample response)
My friend's rule is wrong because the adjacent is not defined by the triangle's orientation on the page; it is defined relative to the marked angle θ. The hypotenuse is fixed — it is always opposite the right angle and always the longest side — but the labels "opposite" and "adjacent" depend entirely on which non-right angle is marked. The opposite is the side that sits directly across from θ; the adjacent is the side touching θ that is not the hypotenuse. If you rotate the same triangle 90°, the physical sides do not change but what counts as "the bottom" does — and so the friend's bottom-side rule would mislabel the triangle. The fix is to ignore orientation entirely: locate the marked angle, locate the hypotenuse, then label the remaining two sides as opposite (across) and adjacent (next to).
Marking: 1 for naming the marked angle as the defining feature, 1 for the hypotenuse being fixed by the right angle, 1 for opposite = across / adjacent = next to (not bottom), 1 for the rotation argument.