Mathematics • Year 10 • Unit 3 • Lesson 7
Mixed Right-angled Triangle Problems — Skill Drill
Build fluency with the Lesson 7 toolkit: exact values for 30°, 45° and 60°; the strategy decision (Pythagoras vs trig vs inverse trig); and the four-step solving pattern (read → draw → choose → check). One step at a time, from a fully worked example to independent practice.
1. I do — fully worked example
Read every step. Each one has a short reason on the right so you can see why, not just what.
Problem. A 12 m rope is tied from the top of a flagpole to a stake in the ground, making a 30° angle with the ground. Find the height of the flagpole using exact values.
Step 1 — Draw and label.
Rope = hypotenuse = 12 m. Angle with ground = 30°. Height = h (opposite the 30° angle).
Reason: the pole is vertical, the ground is horizontal, the rope is the slope. The angle at the ground sits between rope and ground.
Step 2 — Choose the ratio.
Opposite (h) and hypotenuse (12) → use sine (SOH).
Reason: the side I want is opposite the known angle, and the known side is the hypotenuse.
Step 3 — Substitute the exact value sin 30° = 1/2.
sin 30° = h/12 → 1/2 = h/12 → h = 12 × (1/2) = 6 m
Reason: sin 30° = 1/2 exactly — no calculator needed.
Step 4 — Reasonableness check.
At 30°, the opposite side is exactly half the hypotenuse. 6 = 12/2 ✓
Answer: The flagpole is h = 6 m tall.
2. We do — fill in the missing steps
Same structure as Section 1, but with the working faded. Fill in each blank line. 4 marks
Problem. An equilateral triangle has side length 10 cm. Find its perpendicular height using exact values.
Step 1 — Split the equilateral triangle. Drop a perpendicular from one vertex to the opposite side. This bisects the base, creating two ____-____-____ right triangles.
Step 2 — Label one of the right triangles.
Hypotenuse = ______ cm, base = ______ cm, height = h, angle at the base = ______°
Step 3 — Use the exact value for sin 60°.
sin 60° = h / ______ → ______ = h / 10
Step 4 — Solve for h in exact form.
h = 10 × ______ = ______ cm
Step 5 — Approximate check.
h ≈ 5 × 1.732 ≈ ______ cm. Less than 10 (the side) ✓
3. You do — independent practice
Show your working in the space under each problem. The first four are foundation (single tool, exact values where possible). The middle two are standard (decide between tools). The last two are extension (multi-step mixed).
Foundation — single tool, exact values welcome
3.1 Find the value of x: a right triangle has hypotenuse 8 cm and angle 60° between the hypotenuse and the side x (so x is the adjacent side). Use exact values. 1 mark
3.2 A right triangle has hypotenuse 14 cm and angle 45°. Find the side opposite the 45° angle in exact form. 1 mark
3.3 A right triangle has the side opposite the 30° angle equal to 5 cm. Find the hypotenuse (use sin 30° = 1/2). 1 mark
3.4 A right triangle has legs 9 cm and 40 cm. Find the hypotenuse. 1 mark
Standard — choose the strategy
3.5 A square has diagonal 8√2 cm. Find the side length, then verify your answer by computing diagonal = side × √2. 2 marks
3.6 A right triangle has two legs equal to 7 cm and 7√3 cm. Find the hypotenuse, and use a trig ratio to confirm one of the acute angles is 60°. 2 marks
Extension — multi-step mixed problems
3.7 A right triangle has hypotenuse 10 cm and one leg 6 cm. (a) Find the third side using Pythagoras. (b) Find both acute angles to 1 d.p. (c) Verify they sum to 90°. 3 marks
3.8 A 5 m ladder leans against a wall at 60° to the ground. (a) Find the height it reaches up the wall using exact values. (b) Find the horizontal distance from the foot of the ladder to the wall using exact values. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (equilateral, side 10 cm)
Step 1: two 30-60-90 right triangles.
Step 2: Hypotenuse = 10 cm, base = 5 cm, h = h, angle at the base = 60°.
Step 3: sin 60° = h / 10, so √3/2 = h/10.
Step 4: h = 10 × √3/2 = 5√3 cm.
Step 5: h ≈ 5 × 1.732 ≈ 8.66 cm ✓.
3.1 — Hypotenuse 8, angle 60° to side x (adjacent)
cos 60° = x/8 → 1/2 = x/8 → x = 4 cm.
3.2 — Hypotenuse 14, opposite side at 45°
sin 45° = opp/14 → (√2/2) = opp/14 → opp = 14 × √2/2 = 7√2 cm (≈ 9.9 cm).
3.3 — Opposite to 30° is 5 cm
sin 30° = 5/hyp → 1/2 = 5/hyp → hyp = 10 cm.
3.4 — Legs 9 and 40
c² = 81 + 1600 = 1681 → c = 41 cm. (9-40-41 triple.)
3.5 — Square with diagonal 8√2
Diagonal = side × √2, so side = 8√2 / √2 = 8 cm.
Verify: 8 × √2 = 8√2 ✓.
3.6 — Legs 7 and 7√3
c² = 7² + (7√3)² = 49 + 49 × 3 = 49 + 147 = 196 → c = 14 cm.
tan θ = 7√3 / 7 = √3 → θ = tan⁻¹(√3) = 60° ✓.
This is the standard 30-60-90 triangle with sides in ratio 1 : √3 : 2 — scaled by 7.
3.7 — Hypotenuse 10, leg 6
(a) Third side: a² = 100 − 36 = 64 → a = 8 cm (6-8-10 triple).
(b) sin θ = 6/10 = 0.6 → θ ≈ 36.9°. sin φ = 8/10 = 0.8 → φ ≈ 53.1°.
(c) 36.9 + 53.1 = 90° ✓.
3.8 — 5 m ladder at 60°
(a) Wall height = 5 × sin 60° = 5 × (√3/2) = 5√3/2 m (≈ 4.33 m).
(b) Horizontal distance = 5 × cos 60° = 5 × (1/2) = 5/2 = 2.5 m.
Check: (5√3/2)² + (5/2)² = 75/4 + 25/4 = 100/4 = 25 = 5² ✓.