Mathematics • Year 10 • Unit 3 • Lesson 7
Mixed Right-angled Triangle Problems — Mixed Challenge
Pull together every idea from Lesson 7: exact values for 30°, 45° and 60°; strategy choice (Pythagoras, trig, inverse trig); the two special triangles (30-60-90 and 45-45-90); and reasonableness checks. Spot a classmate's plausible mistake, then design your own exact-value problem.
1. Mixed problems — choose the right strategy
Each question uses a different idea from Lesson 7. State which strategy you will use before calculating. Show your working. 3 marks each
1.1 An equilateral triangle has side length 6 cm. Find its perpendicular height using exact values, then convert to a decimal (2 d.p.).
1.2 A square has diagonal 12 cm. Find the side length in exact form, and find the area of the square.
1.3 A 30-60-90 triangle has its shortest side equal to 4 cm. State the other two sides in exact form using the 1 : √3 : 2 ratio.
1.4 A right triangle has legs 11 cm and 60 cm. Find the hypotenuse and identify the Pythagorean triple.
1.5 A flagpole stands on horizontal ground. From a point 18 m from its base, the angle of elevation to the top is 30°. Find the height of the flagpole using exact values.
1.6 A skateboard ramp has a sloping face 2 m long that rises to a deck 1 m above the ground. (a) Find the angle the slope makes with the ground using an exact-value trig ratio. (b) Find the horizontal length of the slope's footprint (exact value).
2. Find the mistake
Another Year 10 student is finding the height of an equilateral triangle with side 8 cm. Their working is below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — height of equilateral triangle, side 8 cm:
Line 1: Drop a perpendicular from one vertex. It splits the base into two pieces of 4 cm each.
Line 2: The right triangle has hypotenuse 8, base 4, and angle at the base of 60°.
Line 3: Use sin 60° = h/8. sin 60° = 1/2, so h = 8 × (1/2) = 4 cm.
Line 4: So the height of the equilateral triangle is 4 cm.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the corrected final answer in exact form.
Stuck? Look at the exact values: sin 30° = 1/2 (not sin 60°). sin 60° = √3/2. Has the student mixed them up?3. Open-ended challenge — design an exact-value problem
This question has many valid answers. Be creative but show every number. 4 marks
3.1 Design a real-world right-angled triangle problem (e.g. ladder, kite, ramp, flagpole shadow, sail) that can be fully solved without a calculator using exact values for 30°, 45° or 60°. Your problem must:
(i) Have a clear context with a labelled diagram.
(ii) Use exactly one of the special angles (30°, 45°, or 60°).
(iii) Produce an exact-value answer (a whole number, a simple fraction, or one involving √2 or √3).
(iv) End with a sanity check confirming the answer is physically reasonable.
Then solve your own problem, writing the exact value and a decimal approximation.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Equilateral, side 6 cm
Split into two 30-60-90 right triangles. Hypotenuse = 6, base = 3, angle at base = 60°.
sin 60° = h/6 → h = 6 × (√3/2) = 3√3 cm ≈ 5.20 cm.
1.2 — Square with diagonal 12 cm
Side = diagonal/√2 = 12/√2 = 12√2/2 = 6√2 cm (≈ 8.49 cm).
Area = side² = (6√2)² = 72 cm².
1.3 — 30-60-90 with shortest side 4
The shortest side is opposite the 30° angle. Ratio 1 : √3 : 2.
Middle side (opp 60°) = 4√3 cm. Hypotenuse = 8 cm.
1.4 — Legs 11 and 60
c² = 121 + 3600 = 3721 → c = 61 cm. Pythagorean triple: 11-60-61.
1.5 — Flagpole, 18 m base, 30° elevation
tan 30° = h/18, and tan 30° = 1/√3.
h = 18 × (1/√3) = 18/√3 = 18√3/3 = 6√3 m ≈ 10.39 m.
1.6 — Skateboard ramp, slope 2 m, height 1 m
(a) sin θ = opposite/hypotenuse = 1/2 → θ = 30°.
(b) Horizontal footprint = 2 × cos 30° = 2 × (√3/2) = √3 m ≈ 1.73 m.
2 — Find the mistake
(a) The mistake is on Line 3.
(b) The student wrote sin 60° = 1/2, but that is the value of sin 30°, not sin 60°. The correct exact value is sin 60° = √3/2. Mixing up the two special angles is the single most common Year 10 error.
(c) Corrected working:
sin 60° = h/8, and sin 60° = √3/2.
h = 8 × (√3/2) = 4√3 cm ≈ 6.93 cm.
Sanity check: the height of an equilateral triangle is less than the side (8 cm), and 6.93 < 8 ✓. With the student's wrong answer of 4 cm, the triangle would be impossibly flat.
3 — Open-ended challenge (sample solution)
Problem. A kite is flying on 60 m of taut string. The string makes a 60° angle with the horizontal ground. Find (a) the height of the kite using an exact value, and (b) the horizontal distance from the child to the point directly below the kite, also as an exact value.
Diagram: right triangle with hypotenuse 60 m (the string), angle 60° at the ground, opposite side = height h, adjacent side = horizontal distance d.
Solution.
(a) sin 60° = h/60 → h = 60 × (√3/2) = 30√3 m ≈ 51.96 m.
(b) cos 60° = d/60 → d = 60 × (1/2) = 30 m.
Sanity check. Pythagoras: (30√3)² + 30² = 2700 + 900 = 3600 = 60² ✓. Both answers fit a 60 m string. 51.96 m is a plausible kite altitude.
Marking: 1 for a clear real-world setup with diagram, 1 for using a special angle (30°, 45°, or 60°), 1 for an exact-value answer involving √2 or √3 (or a simple fraction), 1 for a sanity-check using Pythagoras or real-world reasoning. Full marks for any valid problem and solution meeting all four rules.