Mathematics • Year 10 • Unit 3 • Lesson 15
Sine Rule in the Field — Triangles That Aren't Right-Angled
Apply the sine rule from Lesson 15 to real Australian non-right-angled triangles — a surveyor sighting across the Murray, a yacht bearing problem off Sydney Heads, the angle between two roads at a Newcastle roundabout, and the distance from a lighthouse to two ships. Decide first whether to use the side form (AAS → find side) or the angle form (SSA → find angle), and stay alert to the ambiguous case.
1. Word problems
For each problem: sketch the triangle, label the known sides and angles, decide which form of the sine rule to use, then solve. Give answers to 1 decimal place.
1.1 — Surveying across the Murray. A surveyor at point A wants to know the distance to a tree T directly across the river. She walks 80 m along the south bank to point B. From A she measures ∠TAB = 75° and from B she measures ∠TBA = 65°.
(a) Find ∠ATB.
(b) Use the sine rule to find AT (the distance from A to the tree). 4 marks
1.2 — Yacht off Sydney Heads. A yacht sails from port P on a bearing of 050° for 6 km to point Q, then changes course and sails to point R such that ∠PQR = 110° and ∠QRP = 30°.
(a) Find ∠QPR.
(b) Use the sine rule to find the distance QR. 3 marks
1.3 — Newcastle roundabout. Two roads meet at a roundabout. The first road runs 120 m from the roundabout to a service-station entrance S. The second road runs 90 m from the roundabout to a shop entrance T. The straight-line distance from S to T is 70 m. Use the sine rule to find the angle ∠RST (where R is the roundabout). 3 marks
1.4 — Lighthouse and two ships. From a lighthouse L, two ships A and B are observed at the same moment. The bearing to A is 020° at a distance of 8 km. The bearing to B is 080°. The captain of A reports B as being on a bearing of 110° from her position.
(a) Find ∠ALB and ∠LAB.
(b) Use the sine rule to find the distance LB. 4 marks
1.5 — Ambiguous case in the field. A radar operator at base B sights an aircraft A that is 15 km away. A second tracking station T is 12 km from the aircraft. The angle at the aircraft, ∠BAT, is 30°. Use the sine rule to find the two possible values of ∠ATB (the angle at the second station). Explain which case might apply in a real situation. 3 marks
2. Explain your thinking
Use full sentences. 4 marks
2.1 A friend says: "We can always just use SOH CAH TOA — we don't need the sine rule." Using the language of Lesson 15 (right-angled vs non-right-angled, opposite, the side-angle pairing), explain in 4-6 sentences (i) why SOH CAH TOA is not enough for every triangle, (ii) the one extra piece of information the sine rule needs that SOH CAH TOA does not, and (iii) one real-world setup (your choice — surveying, sailing, sports) where the sine rule is the only sensible tool. Include the sine-rule formula in your explanation.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Murray surveying
(a) ∠ATB = 180° − 75° − 65° = 40°.
(b) AT is opposite ∠ABT = 65°. AB = 80 m is opposite ∠ATB = 40°.
AT / sin 65° = 80 / sin 40°.
AT = (80 × sin 65°) / sin 40° ≈ (80 × 0.9063) / 0.6428 ≈ 112.8 m.
1.2 — Sydney yacht
(a) ∠QPR = 180° − 110° − 30° = 40°.
(b) PQ = 6 km is opposite ∠QRP = 30°. QR is opposite ∠QPR = 40°.
QR / sin 40° = 6 / sin 30°.
QR = (6 × sin 40°) / sin 30° = (6 × 0.6428) / 0.5 ≈ 7.7 km.
1.3 — Newcastle roundabout
(Using ∠TRS = 30° given in the hint.) In △RST: RT = 90 m is opposite ∠RST. ST = 70 m is opposite ∠TRS = 30°.
sin(∠RST) / 90 = sin 30° / 70.
sin(∠RST) = (90 × 0.5) / 70 = 45/70 ≈ 0.6429.
∠RST = sin⁻¹(0.6429) ≈ 40.0°.
(The obtuse alternative 180° − 40° = 140° would force ∠RTS < 10°, geometrically possible but the acute answer is the realistic one for a roundabout.)
1.4 — Lighthouse and two ships
(a) ∠ALB = 80° − 20° = 60°. ∠LAB = 200° − 110° = 90°. So ∠LBA = 180° − 60° − 90° = 30°.
(b) LA = 8 km is opposite ∠LBA = 30°. LB is opposite ∠LAB = 90°.
LB / sin 90° = 8 / sin 30°.
LB = (8 × 1) / 0.5 = 16 km.
Check: △LAB has a right angle at A; in a right triangle with hypotenuse opposite the right angle, LB is the hypotenuse, and LB = LA / sin 30° = 8 / 0.5 = 16 ✓.
1.5 — Radar ambiguous case
sin(∠ATB) / AB = sin(∠BAT) / BT, so sin(∠ATB) / 15 = sin 30° / 12.
sin(∠ATB) = (15 × 0.5) / 12 = 7.5 / 12 = 0.625.
Acute ∠ATB = sin⁻¹(0.625) ≈ 38.7°. Obtuse ∠ATB = 180° − 38.7° ≈ 141.3°.
Check obtuse: 30° + 141.3° = 171.3° < 180° → still a valid triangle.
Real-world: a radar usually has additional information (range gates, doppler, direction-of-arrival) that picks the correct answer. Without that, the operator must consider both possibilities — this is exactly the ambiguous case Lesson 15 warns about.
2.1 — Explain your thinking (sample response)
My friend is wrong: SOH CAH TOA only works in right-angled triangles, because the three ratios are defined as opposite/hypotenuse, adjacent/hypotenuse and opposite/adjacent — all of which assume a right angle. Lots of real triangles are not right-angled (a yacht's three-leg course, a surveyor's triangle across a river, a roundabout with two roads), and for those we need the sine rule: a / sin A = b / sin B = c / sin C. The sine rule needs one extra thing SOH CAH TOA does not — at least one matched pair of a side and its opposite angle (e.g. side a opposite ∠A). A good real example is the Murray-river surveying problem: the surveyor measures two angles and the baseline between her two positions, none of the triangle's angles is 90°, but the sine rule lets her solve for the distance to the tree on the other bank.
Marking: 1 for stating SOH CAH TOA needs a right angle, 1 for the "matched side-angle pair" requirement, 1 for stating the sine-rule formula, 1 for a sensible real-world example.