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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.