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.

Apply · Real-World Maths

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

Stuck? The ladder is the hypotenuse, it's the longest side and opposite the right angle (which sits where the wall meets the ground).

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

Stuck? The string is the hypotenuse (25 m). The kite is 24 m above the sand, that's the side opposite θ.

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

Stuck? Vertical drop = opposite, horizontal distance = adjacent, slope length = hypotenuse. This is a 3-4-5 triangle scaled by 20.

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

Stuck? Adjacent is the horizontal distance (next to θ but not the hypotenuse). Cosine uses adjacent and hypotenuse.

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

Stuck? Revisit lesson § "Exact Values", the 30-60-90 triangle has sides 1, √3 and 2.

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.

Stuck? Revisit lesson § "Spot the Trap", the adjacent is defined relative to the marked angle, not the orientation of the triangle.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers, Do not peek before attempting

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.