Mathematics • Year 10 • Unit 3 • Lesson 2

Finding Unknown Sides — Mixed Challenge

Pull together every idea from Lesson 2: angle + given + unknown, match-the-pair, rearrangement (unknown on top vs unknown on the bottom), DEG mode, exact values for 30°/45°/60°, and rounding. Spot the classic RAD-mode mistake, then design two triangles that share a property.

Master · Mixed Challenge

1. Mixed problems — choose the right tool

Each question pulls on a different idea from Lesson 2. Decide which ratio applies before you start writing. Round to 2 d.p. unless told otherwise. 2-3 marks each

1.1 Hypotenuse 18 m, marked angle 55°. Find the opposite side. 2 marks

1.2 Adjacent 7.5 cm, marked angle 35°. Find the opposite side. 2 marks

1.3 Opposite 12 m, marked angle 48°. Find the hypotenuse. 2 marks

1.4 Without a calculator, find the side opposite the 60° angle in a right-angled triangle with hypotenuse 10 cm. Leave the answer in exact (surd) form. 2 marks

1.5 A right-angled triangle has a 25° angle, and the side adjacent to that angle is 6 m. Find the hypotenuse and the opposite side. 3 marks

1.6 A flagpole at a Brisbane school is 9 m tall. A wire is fixed from the top of the pole to the ground, making an angle of 65° with the ground. Find the length of the wire and how far from the base of the pole it is fixed. 3 marks

Stuck on 1.6? Pole = opposite to the 65° angle. Wire = hypotenuse (sine). Distance to base = adjacent (tangent or cosine).

2. Find the mistake

A Year 10 student tries to find the opposite side in a triangle with hypotenuse 20 and angle 25°. Their working is shown. Exactly one line contains a mistake. Spot it, explain why, and re-do the working correctly. 3 marks

Student's working — find opposite, H = 20, angle = 25°:

Line 1: H = 20, angle = 25°, unknown = O = x.

Line 2: sin 25° = x / 20

Line 3: x = 20 × sin 25°

Line 4: x = 20 × 0.4226 = 8.452 (calc said 25 → 0.4226 in RAD mode)

Line 5: x ≈ 8.45 cm

(a) Which line contains the mistake?

(b) Explain in one or two sentences what the student got wrong and why their final answer is still correct anyway.

(c) State what the correct annotation on Line 4 should be, and what value the calculator should produce for sin 25°.

Stuck? In DEG mode, sin 25° ≈ 0.4226. In RAD mode, sin 25° (treated as 25 radians) ≈ −0.1324. The student's labelling is wrong, but the number 0.4226 only matches DEG mode.

3. Open-ended challenge — design a pair of triangles

This question has many valid answers. Be creative but show every number. 4 marks

3.1 Design two different right-angled triangles, both with one angle of 40°, where:

Triangle A has a hypotenuse of 10 cm.
Triangle B has a hypotenuse of exactly double that.

For each triangle:
(i) Use sine and cosine to find the opposite and adjacent sides (round to 2 d.p.).
(ii) Confirm that the sides of triangle B are exactly double the sides of triangle A.
(iii) Explain in one sentence why this scaling relationship is guaranteed.

Stuck? Triangles with the same angles are similar — every side scales by the same factor. Doubling the hypotenuse must double every other side too.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — sin 55°, H = 18 m

sin 55° = O / 18 → O = 18 × sin 55° = 18 × 0.8192 = 14.7449 ≈ 14.74 m.

1.2 — tan 35°, A = 7.5 cm

tan 35° = O / 7.5 → O = 7.5 × tan 35° = 7.5 × 0.7002 = 5.2514 ≈ 5.25 cm.

1.3 — sin 48°, O = 12 m, find H

sin 48° = 12 / x → x = 12 / sin 48° = 12 / 0.7431 = 16.1483 ≈ 16.15 m.

1.4 — Exact value, sin 60°, H = 10 cm

sin 60° = O / 10 → O = 10 × sin 60° = 10 × (√3 / 2) = 5√3 cm (≈ 8.66 cm).

1.5 — 25°, A = 6 m, find H and O

cos 25° = 6 / H → H = 6 / cos 25° = 6 / 0.9063 = 6.6201 ≈ 6.62 m.
tan 25° = O / 6 → O = 6 × tan 25° = 6 × 0.4663 = 2.7980 ≈ 2.80 m.
Pythagoras check: 6² + 2.80² = 36 + 7.84 = 43.84, √43.84 ≈ 6.62. ✓

1.6 — Flagpole and wire

sin 65° = 9 / wire → wire = 9 / sin 65° = 9 / 0.9063 = 9.9298 ≈ 9.93 m.
tan 65° = 9 / distance → distance = 9 / tan 65° = 9 / 2.1445 = 4.1968 ≈ 4.20 m.
The wire (hypotenuse) is just longer than the 9 m pole, as expected for a 65° angle.

2 — Find the mistake

(a) The mistake is on Line 4 — specifically the annotation "(calc said 25 → 0.4226 in RAD mode)".
(b) The student has mis-labelled the calculator mode. The value 0.4226 only appears in DEG mode; in RAD mode the calculator would interpret 25 as 25 radians and return sin 25 ≈ −0.1324. Their numerical answer (8.45 cm) is still correct, because the value 0.4226 is correct — they just wrote the wrong label.
(c) Line 4 should read: "x = 20 × sin 25° = 20 × 0.4226 = 8.452 (calculator in DEG mode)". And the displayed value should still be sin 25° ≈ 0.4226.
The lesson's "Check and Round" card flags this exact trap — always confirm the screen says DEG (D) before pressing sine.

3 — Open-ended (sample solution)

Triangle A (H = 10 cm, 40°):
O = 10 × sin 40° = 10 × 0.6428 = 6.43 cm.
A = 10 × cos 40° = 10 × 0.7660 = 7.66 cm.

Triangle B (H = 20 cm, 40°):
O = 20 × sin 40° = 20 × 0.6428 = 12.86 cm.
A = 20 × cos 40° = 20 × 0.7660 = 15.32 cm.

Doubling check: 6.43 × 2 = 12.86 ✓ and 7.66 × 2 = 15.32 ✓. Every side in B is exactly double the matching side in A.

Why it's guaranteed: any two right-angled triangles with the same angles are similar, so their sides are in the same ratio. Doubling the hypotenuse forces every other side to double. (This is the "Same angle, same ratio" anchor from Lesson 1.)

Marking: 1 for each triangle's two sides correctly calculated (× 2 = 2 marks), 1 for verifying the doubling, 1 for the similarity explanation.