Trig Ratios for Any Angle: CAST & Reference Angles (Grade 10)

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Grade 10 · Trigonometry

MARKING GUIDELINE

Marks

Duration

50 minutes

Questions

Name:

Class:

Date:

Mark

/ 33

Instructions and Information

Answer ALL the questions in this question paper.
Answer QUESTION 1 by circling the letter (A–D) in the answer grid at the end of that section.
Show ALL calculations clearly.
Show all units where applicable.
Number the answers correctly according to the numbering system used in this question paper.
A non-programmable calculator may be used, unless stated otherwise.
Write neatly and legibly.

Question 1

[10 MARKS]

Four options are given as possible answers to the following questions. Choose the correct answer and circle the letter (A–D) in the grid at the end of this section. If you want to change your choice, put a cross through the wrong letter and circle your new choice.

1.1
305° lies in …
(1)
A)the first quadrant
B)the second quadrant
C)the fourth quadrant✓
D)the third quadrant
Answer: C — 270° < 305° < 360°.
- A — the first quadrant runs only from 0° to 90°
- B — the second quadrant runs from 90° to 180°
- D — the third quadrant stops at 270°
1.2
The reference angle for 210° is …
(1)
A)60°
B)30°✓
C)150°
D)210°
Answer: B — In the third quadrant, β = 210° − 180° = 30°.
- A — that is 270° − 210°, measured from the wrong axis
- C — that is 360° − 210°
- D — a reference angle is always acute
1.3
cos 120° = …
(1)
A)−1/2✓
B)1/2
C)−(√3)/2
D)(√3)/2
Answer: A — Second quadrant, β = 60°, cosine negative: −cos 60° = −1/2.
- B — forgot that cosine is negative in the second quadrant
- C — used sin 60° instead of cos 60°
- D — the wrong ratio and the wrong sign
1.4
sin θ < 0 and cos θ > 0 in …
(1)
A)the first quadrant
B)the second quadrant
C)the third quadrant
D)the fourth quadrant✓
Answer: D — CAST: only cosine is positive in the fourth quadrant.
- A — both are positive there
- B — there sine is positive and cosine negative
- C — there both are negative
1.5
P(−3 ; 4) lies on the terminal arm of θ. Then cos θ = …
(1)
A)4/5
B)−4/3
C)−3/5✓
D)3/5
Answer: C — r = √(9 + 16) = 5, and cos θ = x/r = −3/5.
- A — that is sin θ = y/r
- B — that is tan θ = y/x
- D — dropped the sign of x
1.6
tan 225° = …
(1)
A)−1
B)1✓
C)√3
D)0
Answer: B — Third quadrant, β = 45°, tangent positive: tan 45° = 1.
- A — tangent is POSITIVE in the third quadrant
- C — that is tan 60°, not tan 45°
- D — that is tan 180°
1.7
If sin θ = 12/13 and θ is obtuse, then cos θ = …
(1)
A)−5/13✓
B)5/13
C)−12/5
D)5/12
Answer: A — Obtuse puts θ in the second quadrant: y = 12, r = 13, so x = −5 and cos θ = −5/13.
- B — an obtuse angle is in the second quadrant, where cosine is negative
- C — that is tan θ
- D — that is the reciprocal of tan θ, with the wrong sign
1.8
cosec 30° = …
(1)
A)1/2
B)2/(√3)
C)√3
D)2✓
Answer: D — cosec 30° = 1/(sin 30°) = 1/(1/2) = 2.
- A — that is sin 30° itself
- B — that is sec 30° = 1 ÷ cos 30°
- C — that is cot 30°
1.9
sec θ is undefined when θ = …
(1)
A)0°
B)180°
C)90°✓
D)45°
Answer: C — sec θ = 1/(cos θ), and cos 90° = 0: division by zero is undefined.
- A — cos 0° = 1, so sec 0° = 1
- B — cos 180° = −1, so sec 180° = −1
- D — cos 45° = (√2)/2, so sec 45° = √2
1.10
sin 330° + cos 240° = …
(1)
A)0
B)−1✓
C)1
D)−√3
Answer: B — sin 330° = −sin 30° = −1/2 and cos 240° = −cos 60° = −1/2; the sum is −1.
- A — both terms are negative, so they cannot cancel
- C — sine is negative in the fourth quadrant and cosine negative in the third
- D — used the ratios of 60° and 30° the wrong way round

Answer grid — marking guideline

1.1	A	B	C	D
1.2	A	B	C	D
1.3	A	B	C	D
1.4	A	B	C	D
1.5	A	B	C	D
1.6	A	B	C	D
1.7	A	B	C	D
1.8	A	B	C	D
1.9	A	B	C	D
1.10	A	B	C	D

Question 2

[10 MARKS]

P(−5 ; 12) lies on the terminal arm of θ, as shown in the diagram below.

2.1
Calculate the length of OP.
(2)
OP² = (−5)² + 12² = 169 (1)
OP = 13 (1)
2.2
Write down the values of sin θ, cos θ and tan θ.
(3)
sin θ = 12/13 (1)
cos θ = −5/13 (1)
tan θ = −12/5 (1)
2.3
Determine the value of cosec θ + cot θ.
(3)
cosec θ = 13/12 (1)
cot θ = −5/12 (1)
cosec θ + cot θ = 8/12 = 2/3 (1)
2.4
Explain, without using a calculator, why 90° < θ < 180°.
(2)
x is negative and y is positive (1)
so P lies in the second quadrant, which runs from 90° to 180° (1)

Question 3

[13 MARKS]

Answer the following WITHOUT using a calculator.

3.1
Evaluate (sin 210° × cos 330°)/(tan 135°).
(5)
sin 210° = −sin 30° = −1/2 (1)
cos 330° = cos 30° = (√3)/2 (1)
tan 135° = −tan 45° = −1 (1)
((−1/2)((√3)/2))/−1 = (√3)/4 (2)
3.2
If 13 cos θ + 5 = 0 and 180° < θ < 360°, determine the value of sin θ + tan θ.
(5)
cos θ = −5/13 (1)
cosine is negative, so θ is in the third quadrant (1)
x = −5, r = 13, so y = −12 (1)
sin θ = −12/13 and tan θ = 12/5 (1)
sin θ + tan θ = −60/65 + 156/65 = 96/65 (1)
3.3
Explain why cosec 180° is undefined.
(3)
cosec θ = 1/(sin θ) (1)
sin 180° = 0 (1)
and division by zero is undefined (1)

TOTAL: 33 marks

This question paper consists of 3 questions.