Mathematics Standard • Year 12 • Trigonometry • Lesson 6
Sine Rule Reasoning, Mastery Challenge
Work the rule backwards, resolve an obtuse angle and prove the triangle closes, chain two triangles together, and diagnose the two errors that cost the most marks in this topic.
1. Working backwards
You are given the angles and one side, and asked for another side. 3 marks
Q1.1 In triangle ABC, A = 50°, B = 75° and b = 24. Find a, to 2 decimal places, and explain in one sentence why a must be shorter than b.
2. The obtuse case, with proof
Part (c) is where the marks separate. 4 marks
Q2.1 In triangle ABC, a = 9, A = 35° and b = 14, and B is known to be obtuse.
(a) Find the value the calculator returns for the inverse sine. (b) State B, to the nearest minute. (c) Show that this triangle actually closes, and explain what you would conclude if it did not.
3. Two triangles chained
The answer to the first triangle is the input to the second. 4 marks
Q3.1 In triangle ABD, angle A = 61°, angle D = 72° and BD = 120 m. Find AC, where C lies on AD such that triangle ABC has the same angle at A and angle ABC = 47°. Start by finding AB, then work in the second triangle. Give AC to 2 decimal places.
4. Find the error
Two students, two different mistakes. 5 marks
Q4.1 Asked to find b when A = 62°, B = 73° and a = 45, a student writes "b = 45 sin62° / sin73° = 41.55". Explain the error and give the correct answer.
Q4.2 Asked for an obtuse angle B, a student computes sin B = 0.8 and writes "B = 53°08′". Explain what has been missed and how to correct it in one step.
5. Why the ambiguity exists
A reasoning question. A number on its own will not score. 3 marks
Q5.1 Explain why a calculator cannot decide on its own whether an angle found by the sine rule is acute or obtuse, and state what extra information a question must supply for the obtuse answer to be the right one.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1, Working backwards
b = 24 pairs with B = 75°, and a pairs with A = 50° [1]. So a = 24 sin50° / sin75° = 24 × 0.766044 / 0.965926 = 19.03 [1].
a must be shorter than b because 50° is smaller than 75°, and in any triangle the bigger angle faces the bigger side [1]. That sense check catches an inverted fraction instantly.
Q2.1, The obtuse case
(a) sin B = 14 sin35° / 9 = 14 × 0.573576 / 9 = 0.892229, and the calculator returns 63.15° [1].
(b) B is obtuse, so B = 180° − 63.1548° = 116.8452°. In minutes, 0.8452 × 60 = 50.7, giving 116° 51′ [2].
(c) A + B = 35° + 116.85° = 151.85°, which is less than 180°, so a third angle of about 28.15° exists and the triangle closes [1]. If the sum had exceeded 180° there would be no such triangle, and the correct conclusion would be that B cannot be obtuse for these measurements, whatever the question asserted.
Q3.1, Two triangles chained
First triangle ABD. Angle ABD = 180° − 61° − 72° = 47°. BD = 120 pairs with A = 61°, and AB pairs with D = 72°, so AB = 120 sin72° / sin61° = 120 × 0.951057 / 0.874620 = 130.49 m [2].
Second triangle ABC. Angle A = 61° and angle ABC = 47°, so angle ACB = 180° − 61° − 47° = 72°. AB = 130.49 pairs with C = 72°, and AC pairs with B = 47° [1], so AC = 130.49 sin47° / sin72° = 130.49 × 0.731354 / 0.951057 = 100.34 m [1].
Carry the full value of AB through, not the rounded 130.49, if your calculator allows it. Rounding at every stage is how a two-triangle answer drifts.
Q4.1, The inverted fraction
The student has paired the sines with the wrong sides. The unknown b sits opposite B = 73°, so B's sine belongs on top: b = 45 sin73° / sin62°, not the other way round [1].
Correctly, b = 45 × 0.956305 / 0.882948 = 48.74 [1]. The sense check exposes it without any arithmetic: 73° is larger than 62°, so b must be longer than a = 45, and the student's 41.55 is shorter.
Q4.2, Stopping at the acute value
The student has taken the inverse sine and stopped [1]. Because sinθ = sin(180° − θ), the inverse sine always returns the acute solution, and the question asked for the obtuse one [1].
One step fixes it: B = 180° − 53°08′ = 126°52′ [1]. Then check the three angles still sum to less than 180°.
Q5.1, Why the ambiguity exists
Sine is positive in both the first and second quadrants, and sinθ = sin(180° − θ) exactly [1]. So a single sine value corresponds to two possible angles, one acute and one obtuse, and the inverse sine function must return exactly one value, which is always the acute one [1].
For the obtuse answer to be correct, the question must state that the angle is obtuse, or supply information that forces it, such as telling you the angle is the largest in the triangle when the side facing it is the longest [1].