Sine, Cosine & Area Rules (Grade 11) — Worksheet & Memo

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Grade 11 · Trigonometry

Sine, Cosine & Area Rules (Grade 11)

MARKING GUIDELINE

Marks

Duration

1 hour 15 minutes

Questions

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/ 50

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
The SINE rule is the one to use when you know …
(1)
A)two angles and a side✓
B)two sides and the angle BETWEEN them
C)all three sides
D)all three angles
Answer: A — The sine rule pairs each side with the angle OPPOSITE it, so it needs one complete pair.
- B — that is the cosine rule's case
- C — that is also the cosine rule's case
- D — three angles fix the shape but no side length
1.2
The COSINE rule a² = b² + c² − 2bc cos A is used when …
(1)
A)two sides and the INCLUDED angle are known✓
B)two angles and a side are known
C)only one side is known
D)the triangle is right-angled
Answer: A — The cosine rule needs the angle BETWEEN the two known sides, or else all three sides.
- B — that is the sine rule's case
- C — one side is never enough
- D — for a right-angled triangle, Pythagoras and SOH-CAH-TOA are simpler
1.3
In △ABC, A = 40°, B = 60° and a = 10. Then b = …
(1)
A)13,47✓
B)7,42
C)8,66
D)15,32
Answer: A — b ÷ sin B = a ÷ sin A, so b = 10 sin 60° ÷ sin 40° = 13,47.
- B — inverted the ratio: that is 10 sin 40° ÷ sin 60°
- C — used 10 sin 60° without dividing by sin 40°
- D — used the third angle, 80°, in place of 60°
1.4
In △PQR, p = 5, q = 7 and R = 60°. Then r = …
(1)
A)6,24✓
B)39
C)10,44
D)2
Answer: A — r² = 5² + 7² − 2(5)(7)cos 60° = 74 − 35 = 39, so r = √39 = 6,24.
- B — forgot to take the square root of 39
- C — ADDED the 35 instead of subtracting it
- D — used 7 − 5 and ignored the cosine rule altogether
1.5
The AREA rule states that the area of a triangle is …
(1)
A)12ab sin C✓
B)12ab cos C
C)ab sin C
D)12 × base × height only
Answer: A — The area rule uses two sides and the SINE of the angle between them.
- B — the area rule uses the SINE of the included angle, not its cosine
- C — the ½ has been left off
- D — that formula needs a perpendicular height, which a general triangle rarely gives
1.6
Find the area of △ABC if b = 8, c = 6 and A = 30°.
(1)
A)12✓
B)24
C)20,78
D)48
Answer: A — Area = ½(8)(6)sin 30° = 24 × 0,5 = 12.
- B — left off the sin 30° factor
- C — used cos 30° instead of sin 30°
- D — left off both the ½ and the sine
1.7
In △ABC, a = 9, b = 7 and c = 5. Then cos A = …
(1)
A)−0,1✓
B)0,1
C)−7
D)0,83
Answer: A — cos A = (7² + 5² − 9²) ÷ (2 × 7 × 5) = −7 ÷ 70 = −0,1.
- B — dropped the minus — 49 + 25 is LESS than 81
- C — forgot to divide by 2bc
- D — used a² + b² − c², which finds cos C instead
1.8
Hence, in that triangle, ∠A = …
(1)
A)95,74°✓
B)84,26°
C)5,74°
D)1,67°
Answer: A — A = cos⁻¹(−0,1) = 95,74°, which is obtuse because the cosine is negative.
- B — that is cos⁻¹(+0,1) — the minus was dropped
- C — subtracted 90° from the correct answer
- D — the calculator was in RADIAN mode
1.9
The AMBIGUOUS case of the sine rule can arise when …
(1)
A)two sides and a NON-included angle are given✓
B)all three sides are given
C)two angles and a side are given
D)one of the angles is a right angle
Answer: A — The unknown angle may be acute or obtuse, and both can fit the given lengths.
- B — three sides fix the triangle completely
- C — two angles fix the third, so the triangle is determined
- D — a right angle leaves no ambiguity
1.10
In △ABC, if C = 90° the cosine rule becomes …
(1)
A)c² = a² + b²✓
B)c² = a² − b²
C)c² = a² + b² − 2ab
D)c = a + b
Answer: A — cos 90° = 0, so the −2ab cos C term vanishes and Pythagoras is what is left.
- B — the term that vanishes is −2ab cos C, not the + b²
- C — cos 90° = 0, so the whole 2ab term goes, not just the cosine
- D — the theorem relates the SQUARES of the sides, not the sides

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

[12 MARKS]

In the diagram below, △ABC is drawn with BÂC = α, AB̂C = β and AB = c.

2.1
Write down the size of AĈB in terms of α and β.
(1)
AĈB = 180° − α − β (1)
2.2
Show that BC = c · sin αsin(α + β).
(4)
By the sine rule: BCsin Â = ABsin Ĉ (1)
BC = c · sin αsin(180° − α − β) (1)
sin(180° − θ) = sin θ (1)
∴ BC = c · sin αsin(α + β) (1)
2.3
Given that α = 52°, β = 61° and c = 48 m, calculate the length of BC, correct to TWO decimal places.
(3)
BC = 48 sin 52°sin 113° (2)
= 41,09 m (1)
2.4
Hence calculate the area of △ABC, correct to TWO decimal places.
(4)
Area = ½ · AB · BC · sin B̂ (the area rule, with the INCLUDED angle) (1)
= ½ × 48 × 41,09 × sin 61° (2)
= 862,54 m² (1)

Question 3

[15 MARKS]

In the diagram below, ABCD is a quadrilateral with the diagonal AC drawn. AB̂C = 106°, BÂC = 31°, AĈD = 46°, AC = 4,6 cm and CD = 10 cm. BC = x. The diagram is not drawn to scale.

3.1
Calculate the length of x, correct to TWO decimal places.
(3)
In △ABC: xsin 31° = 4,6sin 106° (the sine rule) (1)
x = 4,6 sin 31°sin 106° (1)
= 2,46 cm (1)
3.2
Calculate the area of △ABC, correct to TWO decimal places.
(4)
AĈB = 180° − 106° − 31° = 43° (1)
Area = ½ × AC × x × sin(AĈB) (1)
= ½ × 4,6 × 2,46 × sin 43° (1)
= 3,87 cm² (1)
3.3
Calculate the length of AD, correct to TWO decimal places.
(4)
In △ACD: AD² = AC² + CD² − 2(AC)(CD) cos(AĈD) (the cosine rule) (1)
= 4,6² + 10² − 2(4,6)(10) cos 46° (1)
= 57,25 (1)
AD = 7,57 cm (1)
3.4
Hence calculate the area of quadrilateral ABCD, correct to TWO decimal places.
(4)
Area △ACD = ½ × 4,6 × 10 × sin 46° (1)
= 16,54 cm² (1)
Area ABCD = 3,87 + 16,54 (1)
= 20,41 cm² (1)

Question 4

[13 MARKS]

VABC is a pyramid with V the apex and △ABC its horizontal base. BÂC = 110°, AB̂C = 40° and BC = 6 m. The perpendicular height of the pyramid is 8 m. [Volume of a pyramid = ⅓ × area of the base × perpendicular height]

4.1
Write down the size of AĈB, and hence calculate the length of AB, correct to TWO decimal places.
(4)
AĈB = 180° − 110° − 40° = 30° (1)
ABsin 30° = 6sin 110° (the sine rule) (1)
AB = 6 sin 30°sin 110° (1)
= 3,19 m (1)
4.2
Hence calculate the area of the base △ABC, correct to TWO decimal places.
(3)
Area = ½ × AB × BC × sin(AB̂C) (1)
= ½ × 3,19 × 6 × sin 40° (1)
= 6,16 m² (1)
4.3
Hence calculate the volume of the pyramid, correct to TWO decimal places.
(3)
V = ⅓ × 6,16 × 8 (2)
= 16,42 m³ (1)
4.4
Explain why the AREA RULE had to be used for the base rather than ½ × base × perpendicular height.
(3)
The perpendicular height of △ABC is not given (1)
but two sides and the angle BETWEEN them are known (1)
which is exactly what the area rule needs (1)

TOTAL: 50 marks

This question paper consists of 4 questions.