Hint
Use the 3–4–5 reference triangle and track the quadrant of each related angle.
Worked solution
(a)
Acuteness gives cos α=√(1−9/25)=4/5. Hence tan α=(3/5)/(4/5)=3/4.
Answer: \(4/5\) and \(3/4\).
(b)
β is in quadrant III with reference angle α, so both sine and cosine are negative: −3/5 and −4/5.
Answer: \(-3/5\) and \(-4/5\).
(c)
Tangent has period π, so tan β=3/4. The supplementary angle has sine 3/5. Their sum is 3/4+3/5=27/20.
Answer: \(27/20\).
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.
(a) · 2 marks
- 1 mark: Give cos α=4/5 with the acute sign.
- 1 mark: Give tan α=3/4.
Alternative methods: A labelled reference triangle is accepted.
Common error: Using the negative square root for an acute angle.
(b) · 2 marks
- 1 mark: Give sin β=−3/5.
- 1 mark: Give cos β=−4/5.
Alternative methods: A unit-circle half-turn argument is accepted.
Common error: Keeping the quadrant-I signs after a half-turn.
(c) · 3 marks
- 1 mark: Obtain tan β=3/4.
- 1 mark: Obtain sin(π−α)=3/5.
- 1 mark: Add to give 27/20.
Alternative methods: A correct quadrant/reference-angle argument for both terms is accepted.
Common error: Assuming the sum of two different ratios must itself lie between −1 and 1.
Total: 7 marks.