Download MCR 3U Trigonometry Test 7

MCR 3U Functions Grade 11
Trigonometry TEST 7
Knowledge and Understanding
1. Determine the exact value of each expression. [2]
a) csc 330°
Solution
csc 330° = csc (360° – 30°) = – csc 30° = – 2.
b) cot 240°
Solution
cot 240° = cot (180° + 60°) = cot 60° =
7
1
√3
=
√3
.
3
2. If sec 𝜃 = – and 𝜃 is in the third quadrant, find the other five
3
trigonometric ratios. Include a diagram. [3]
Solution
𝑟
7
sec 𝜃 = = – ⟹ x = – 3, r = 7.
𝑥
3
2
2
2
x + y = r ⟹ (– 3)2 + y2 = 72 ⟹ 9 + y2 = 49 ⟹ y2 = 40.
Since 𝜃 is in the third quadrant, we get: y = – 2√10.
𝑦
2√10
𝑟
𝑥
3
7
𝑟
𝑦
7
2√10
𝑥
𝑟
3
𝑦
2√10
3√10
sin θ = = –
,
cos θ = = – ,
tan θ = =
csc θ = = –
cot θ =
3
2√10
.
7
=
=–
20
.
7√10
20
,
3. Given that tan 𝜃 = – 0.7002, determine all possible values of 𝜃, to the
nearest degree, if 0° ≤ 𝜃 ≤ 360°. [2]
Solution
Related angle 𝛼 = tan – 1 (0.7002) ≈ 35°.
tan θ < 0 ∴ θ lies in the II or IV quadrants.
II Quadrant
IV Quadrant
o
o
o
o
θ = 180 – 𝛼 = 180 – 35 = 145 , θ = 360o – 𝛼 = 360o – 35o = 325o.
Answer: θ = 145o or 325o.
4. In ∆ABC, ∠A = 51°, a = 12 cm, and b = 15 cm. Solve the triangle.
Include a diagram. Round all angles to the nearest degree and sides to
the nearest tenth of a centimetre. [4]
ℎ
From ∆ ACH: sin 51o = ⟹
15
o
h = 15 sin 51 ≈ 11.66 cm
Since b = AC = 15 cm, a = 12 cm,
we have: h < a < b ∴ 2 triangles.
From ∆ ABC by Sine Law:
𝐵𝐶
𝐴𝐶
12
15
=
⟹
=
⟹
sin ∠𝐴
sin ∠𝐵
sin ∠B =
sin 51°
15 sin 51o
12
sin ∠𝐵
≈ 0.9714.
∠B = sin– 1 (0.9714) ≈ 76o or ∠B = 180o – 76o = 104o.
Case 1: ∠AB1C = 76o, ∠ACB1 = 180o – 51o – 76o = 53o.
From ∆ ACB1 by Sine Law:
𝐴𝐵1
sin ∠𝐴𝐶𝐵1
=
𝐶𝐵1
sin ∠𝐴
⟹
𝐴𝐵1
sin 530
=
12
sin 51o
⟹ AB1=
12 sin 53o
sin 510
= 12.3 cm
Case 2: ∠AB2C = 104o, ∠ACB2 = 180o – 51o – 104o = 25o.
From ∆ ACB1 by Sine Law:
𝐴𝐵2
sin ∠𝐴𝐶𝐵2
=
𝐶𝐵2
sin ∠𝐴
⟹
𝐴𝐵2
sin 25
0 =
Answer: 12.3 cm, 6. 5 cm.
12
o ⟹ AB1=
sin 51
12 sin 25o
sin 510
= 6.5 cm
Application
1. From a position some distance away from the base of a flagpole,
Justin estimates that the pole is 4.25 m tall at an angle of elevation of
35°. If Justin is 1.65 m tall, use a reciprocal trigonometric ratio to
calculate how far he is from the base of flagpole, to the nearest
hundredth of a meter. Include a diagram. [3]
Solution
1) FC = FB – BC = 4.25 – 1.65 = 2.6 m.
𝐽𝐶
2) From ∆ JFC: cot ∠FJC = ⟹
JC = FC cot ∠FJC =
𝐹𝐶
𝐹𝐶
tan ∠𝐹𝐽𝐶
=
2.6
tan 35°
≈ 3.71 m
2. From a window in a building, the angle of depression to a parked car
is 32°. From a window that is 12 m lower, the angle of depression to the
parked car is 21°. How far is the parked car from the base of the
building, to the nearest metre? Include a diagram. [3]
Solution
1) ∠ ADB = 21°, ∠ ADC = 32°
(as alternating to the given angles)
2) ∠BDC = ∠ ADC – ∠ ADB
= 32° – 21° = 11°.
3) ∠BCD = 90° – 32° = 58°.
4) From ∆ BCD by Sine Law,
𝐵𝐷
𝐵𝐶
𝐵𝐷
=
⟹
=
sin ∠𝐵𝐶𝐷
sin ∠𝐵𝐷𝐶
sin 58°
5) From ∆ ABD, cos ∠ ADB =
12
sin 11°
𝐴𝐷
𝐵𝐷
⟹ BD =
12 sin 58°
sin 11°
≈ 53.334 m
⟹ AD = BD cos ∠ ADB
⟹ AD = 53.334 cos 21° ≈ 49.8 m
3. Given ∆PQR, QR = 5.6 m and S is the midpoint of QR. Determine
PQ, to the nearest tenth, if ∠PSQ = 33° and ∠ PRQ = 22°. Include a
diagram. [3]
Solution
1) By Exterior Angle Theorem,
∠SPR = ∠PSQ – ∠ PRQ = 33° – 22° = 11°.
2) From ∆ PRS by Sine Law,
𝑃𝑆
sin ∠ 𝑃𝑅𝑆
=
𝑅𝑆
sin∠𝑆𝑃𝑅
⟹
𝑃𝑆
sin 22°
=
2.8
sin 11°
⟹ PS =
3) From ∆ PQS by Cosine Law,
PQ2 = QS2 + PS2 – 2 QS ∙ PS cos ∠ PSQ
PQ2 = 2.82 + 5.52 – 2 ∙ 2.8 ∙ 5.5 cos 33°
PQ2 = 12.2589 ⟹ PQ ≈ 3.5 m.
2.8 sin 22°
sin 11°
≈ 5.5 m.