Hint
Pythagoras gives the magnitude of sine. The stated quadrant determines its sign.
Worked solution
(a)
sin²θ=1−25/169=144/169. In quadrant II sine is positive, so sin θ=12/13.
Answer: \(12/13\).
(b)
tan θ=sin θ/cos θ=(12/13)/(−5/13)=−12/5.
Answer: \(-12/5\).
(c)
The numerator is 1/13 and the denominator is 25/169, so the quotient is (1/13)(169/25)=13/25. The denominator is nonzero. Equivalently the quotient is 1/(1+sin θ).
Answer: \(13/25\).
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: Obtain sin²θ=144/169 or the corresponding reference-triangle length.
- 1 mark: Choose the positive value 12/13 using the quadrant.
Alternative methods: A 5–12–13 reference triangle with a correct sign argument is accepted.
Common error: Giving both square-root signs despite the stated quadrant.
(b) · 2 marks
- 1 mark: Use sine divided by cosine.
- 1 mark: Give −12/5.
Alternative methods: A correct reference-triangle ratio with the quadrant sign is accepted.
Common error: Using cosine divided by sine.
(c) · 3 marks
- 1 mark: Obtain the correct numerator and squared denominator.
- 1 mark: Divide the fractions correctly or use a valid identity.
- 1 mark: Give the exact result 13/25.
Alternative methods: Rationalising with the Pythagorean identity is accepted on this valid domain.
Common error: Squaring the denominator ratio incorrectly or changing its sign.
Total: 7 marks.