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Sine Ratio Introduction to Trigonometric Ratios The figure below shows a right-angled triangle ABC, where B = and C = 90. A hypotenuse θ B adjacent side of opposite side of C AB is called the hypotenuse; BC is called the adjacent side of ; AC is called the opposite side of . In fact, thethe size of has certain relationship Consider right-angled PQR PQ PQ among QR between ratios below. Isthe there any relationship and , PR . PR QR , PQ,ratios QR and These arePR? known as trigonometric ratios. P θ Q R 2 + QR2 = PR2… IHow onlydoes knowthe thatsize PQof relate to the sides of the triangle? Consider threeWhat right-angled Completethe thefollowing table below. do you triangles. observe? 4 2 1 A Triangle opposite side hypotenuse 2 B 30 3 6 C 30 30 A B C 1 2 1 2 = 2 4 1 3 = 2 6 For a right-angled triangle with a given acute angle , opposite side is a constant. hypotenuse Concept of Sine Ratio The sine ratio of an acute angle is defined as below: opposite side sin θ hypotenuse hypotenuse θ opposite side of For a right-angled triangle with a given acute angle , the sine ratio of is a constant. For example, 2 30 sin 30 = 6 4 1 2 30 1 2 3 = = 2 4 6 30 3 In ABC, C 90, AB 13 and AC 5. A 13 B AC AC sin B sin B AB AB 5 5 13 13 5 C AC is the opposite side of B, and AB is the hypotenuse. Follow-up question 1 In the following figures, find sin θ. (a) (b) 15 12 17 θ θ 20 16 8 Solution 15 (a) sin θ 17 (b) 16 sin θ 20 4 5 Example 1 In △PQR, ∠P 90, PQ 6, QR 10 and RP 8. Find the values of (a) sin ∠Q, (b) sin ∠R. (Give your answers in fractions.) Solution PR (a) sin Q QR 8 10 4 5 PQ (b) sin R QR 6 10 3 5 Finding Sine Ratio Using Calculators Find sin for a given angle 1. Make sure that the calculator is set in degree mode. Degree mode is usually denoted by the key DEG or D on calculators. 2. Use the key sin on a calculator to find the value of sin . For example, the value of sin 30 can be obtained by keying: sin 30 EXE The answer is 0.5. Follow-up question 2 By using a calculator, find the values of the following expressions correct to 4 significant figures. (a) sin 43 – sin 28 (b) 2 sin 11 Solution (a) sin 43 – sin 28 = 0.2125 (b) 2 sin 11 = 0.3816 sin 43 = 0.681 99…, sin 28 = 0.469 47… (cor. to 4 d.p.) sin 11 = 0.190 80… (cor. to 4 d.p.) Example 2 By using a calculator, find the values of the following expressions correct to 4 decimal places. (a) sin 66 Solution (a) (b) sin 32.48 Keying sequence Display sin 66 EXE 0.913545457 sin 66 0.9135 (cor. to 4 d.p.) (b) Keying sequence Display sin 32.48 EXE 0.537005176 sin 32.48 0.5370 (cor. to 4 d.p.) Example 3 (a) By using a calculator, find the value of sin 34 + sin 26 (b) sin 60 correct to 3 significant figures. From the result obtained in (a), is sin 34 + sin 26 equal to sin (34 + 26)? Solution (a) Keying sequence Display sin 34 + sin 26 –sin 60 EXE 0.131538646 sin 34 sin 26 sin 60 0.132 (cor. to 3 sig. fig.) (b) ∵ sin 34 sin 26 sin 60 0 sin 34 sin 26 sin 60 sin 34 sin 26 sin (34 26) Find for a given value of sin In degree mode, use the keys SHIFT and sin to find the corresponding acute angle . For example, given that sin = 0.5, can be obtained by keying: SHIFT sin 0.5 EXE The answer is 30, i.e. = 30. Follow-up question 3 Find the acute angle in each of the following using a calculator. (Give your answers correct to 3 significant figures.) (a) sin = 0.22 (b) sin = sin 68 – sin 40 Solution (a) sin = 0.22 = 12.7 (cor. to 3 sig. fig.) (b) sin = sin 68 – sin 40 sin 68 = 0.927 18…, sin 40 = 0.642 78… = 0.2844… = 16.5 (cor. to 3 sig. fig.) Example 4 Find the acute angles in the following using a calculator. (a) sin 0.62, correct to the nearest degree. 1 sin 35, correct to the nearest 0.1. 5 (b) sin (c) 7 sin 3, correct to 3 significant figures. Solution (a) Keying sequence Display SHIFT sin 0.62 EXE 38.31613447 sin 0.62 38 (cor. to the nearest degree) (b) Keying sequence Display SHIFT sin ( 1 5 sin 35 ) EXE 6.587203533 1 sin sin 35 5 6.6 (cor. to the nearest 0.1) (c) 7 sin 3 3 sin 7 Keying sequence Display SHIFT sin ( 3 7 ) EXE 25.37693352 3 sin 7 25.4 (cor. to 3 sig. fig.) Using Sine Ratio to Find Unknowns in Right-Angled Triangles We can use the sine ratio to solve problems involving right-angled triangles. In ABC, C = 90, B = 55 and AB = 8 m. A Find AC correct to 2 decimal places. AC sin B AB AC sin 55 8m AC 8 sin 55 m 6.55 m (cor. (cor. 2 d.p.) to 2tod.p.) 8m 55 B C In PQR, R = 90, PQ = 9 m and PR = 7 m. P Find Q correct to 2 decimal places. AC sin B AC PR sin B Q AB AB PQ AC sin 55 7 m AC 8 m sin 55 Q m 55 m AC 89 sin m AC sin 55 m to 2 d.p.) Q 8 51.06 (cor. 6.55 m (cor. to 2 d.p.) 7m 9m R Q Follow-up question 4 In C ABC, B 75, C 90 and AC 7 cm. Find AB correct to 2 decimal places. Solution AC AC sin B sin B AB AB m 77cm cm sin sin75 75 AB AB AB 77 AB cm m AB cm sin sin75 75 7.25 (cor. to2d.p.) 2d.p.) d.p.) m (cor. 7.25cm cm (cor. toto2 7 cm A 75 B Example 5 In △ABC, ∠B 90, ∠C 42 and AC 5 cm. Find the length of AB correct to 1 decimal place. Solution ∵ AB sin C AC ∴ sin 42 AB 5 cm AB 5 sin 42 cm 3.3 cm (cor. to 1 d.p.) Example 6 In △ABC, ∠B 38, ∠C 90 and AC 15 cm. Find the length of AB correct to 1 decimal place. Solution ∵ ∴ AC sin B AB 15 cm sin 38 AB 15 AB cm sin 38 24.4 cm (cor. to 1 d.p.) Example 7 In △ABC, ∠C 90, AB 17 cm and BC 9 cm. Find ∠A correct to the nearest degree. Solution ∵ ∴ BC sin A AB 9 cm 17 cm A 32 (cor. to the nearest degree) Cosine Ratio Concept of Cosine Ratio The cosine ratio of an acute angle is defined as below: adjacent side cos θ hypotenuse hypotenuse θ adjacent side of For a right-angled triangle with a given acute angle , the cosine ratio of is a constant. For example, 6 4 2 60 1 cos 60 = 60 60 2 1 2 3 = = 2 4 6 3 In ABC, C = 90, AB = 5.2 and AC = 2. A 5.2 B AC AC AC AB AB AB 2 2 2 5.2 5.2 5.2 5 5 5 13 13 13 cos A cos cos A A 2 C AC is the adjacent side of A, and AB is the hypotenuse. Follow-up question 5 In the following figures, find cos θ. (a) (b) 6 θ 3 5 8 θ 10 4 Solution (a) cos θ 3 5 (b) 8 cos θ 10 4 5 Example 8 In △PQR, ∠P 90, PQ 20, PR 21 and RQ 29. Find the values of (a) cos ∠Q, (b) cos ∠R. (Give your answers in fractions.) Solution PQ (a) cos Q QR 20 29 PR QR 21 29 (b) cos R Finding Cosine Ratio Using Calculators Find cos for a given angle In degree mode, use the key cos to find the value of cos . For example, the value of cos 30 can be obtained by keying: cos 30 EXE The answer is 0.8660… Follow-up question 6 By using a calculator, find the values of the following expressions correct to 3 significant figures. cos 40 cos 75 (a) 5 cos 29 (b) 2 Solution (a) 5 cos 29 = 4.37 (b) cos 29 = 0.874 61… (cor. to 3 sig. fig.) cos 40 cos 75 cos 40 = 0.766 04…, cos 75 = 0.258 81… 2 = 0.512 (cor. to 3 sig. fig.) Example 9 By using a calculator, find the values of the following expressions correct to 4 decimal places. (a) cos 12.3 (c) cos 10 cos 72 5 (b) 7 cos 81 5 Solution (a) Keying sequence Display cos 12.3 EXE 0.977045574 cos 12.3 0.9770 (cor. to 4 d.p.) (b) Keying sequence Display ( 7 5 ) cos 81 EXE 0.219008251 7 cos 81 0.2190 (cor. to 4 d.p.) 5 (c) Keying sequence Display cos 72 – cos 10 5 EXE 0.112055443 cos 10 cos 72 0.1121 (cor. to 4 d.p.) 5 Find from a given value of cos In degree mode, use the keys SHIFT and cos to find the corresponding acute angle . For example, given that cos = 0.5, can be obtained by keying SHIFT cos 0.5 EXE The answer is 60, i.e. = 60. Follow-up question 7 Find the acute angle in each of the following using a calculator. (Give your answers correct to 3 significant figures.) (a) cos θ 0.474 cos 24 (b) cos θ 2 Solution (a) cos θ 0.474 1. cos θ 0.474 1. cos θ 0.474 61.7 (cor. sig. fig.) degree) θθ 62 (cor. to to the3 nearest cos 24 24 cos 2. cos cos θθ 2. cos 24 = 0.913 54… (b) 2 2 62.82 (cor. 4 sig. fig.) θθ 62.8 (cor. to to 3 sig. fig.) Example 10 Find the acute angles in the following using a calculator. (a) (b) cos 0.583, correct to the nearest degree. cos 2 cos 75, correct to the nearest 0.1. (c) 12 cos 5, correct to 3 significant figures. Solution (a) Keying sequence Display SHIFT cos 0.583 EXE 54.33817552 cos 0.583 54 (cor. to the nearest degree) (b) Keying sequence Display SHIFT cos ( 2 cos 75 ) EXE 58.8260478 cos 2 cos 75 58.8 (cor. to the nearest 0.1) (c) 12 cos 5 cos 5 12 Keying sequence Display SHIFT cos ( 5 12 ) EXE 65.37568165 5 cos 12 65.4 (cor. to 3 sig. fig.) Using Cosine Ratio to Find Unknowns in Right-Angled Triangles We can use the cosine ratio to solve problems involving right-angled triangles. In ABC, C = 90, B = 55 and AB = 8 m. Find BC correct to 2 decimal places. BC BC 8m cos cos B B BC AB cos B AB BC AB 55 cos 55 40 BC B 9 m 8 cos 40 m 40 m BC 99 cos BC 98 cos cos 55 m BC 40 m 6.89 m (cor. to 2 d.p.) 6.89 m (cor. to 2 4.59 m (cor. to 2 d.p.)d.p.) A C In PQR, R = 90, PQ = 9 m and QR = 7 m. Find Q correct to 2 decimal places. P 9m BC cos B QR cos Q AB BC cos B PQ BC cos 40 AB BC 7m 9 cos 40Q cos m m 9cos m 40 m BC 99 BC 9 cos 40 m Q 38.94 (cor. to 2 d.p.) (cor. 6.89 6.89 m m (cor. to to 2 2 d.p.) d.p.) Q 7m R Follow-up question 8 In ABC, C 90, AB 4 cm and BC 3.5 cm. B Find B correct to 2 decimal places. 3.5 cm C 4 cm Solution BC BC cos BB BC cos cos B AB AB AB BC BC cos35 35 3.5 cm cos cos B 44 cm cm 4 cm BC 44 cos cos 35 35 cm BC BC 4 cos 35 cm B 3.28 28.96cm (cor.(cor. to 2to d.p.) 2 d.p.) 3.28 cm (cor. to 2 d.p.) A Example 11 In △DEF, ∠D 90, ∠E = 62 and EF = 8 cm. Find the length of DE correct to 1 decimal place. Example 12 In △PQR, ∠P 36, ∠Q 90 and PQ 10 cm. Find the length of PR correct to 1 decimal place. Example 13 In △PQR, ∠R 90, PQ = 22 cm and QR =18 cm. Find ∠Q correct to the nearest 0.01. Example 11 In △DEF, ∠D 90, ∠E = 62 and EF = 8 cm. Find the length of DE correct to 1 decimal place. Solution ∵ DE cos E EF ∴ cos 62 DE 8 cm DE 8 cos 62 cm 3.8 cm (cor. to 1 d.p.) Example 12 In △PQR, ∠P 36, ∠Q 90 and PQ 10 cm. Find the length of PR correct to 1 decimal place. Solution ∵ ∴ PQ cos P PR 10 cm cos 36 PR 10 PR cm cos 36 12.4 cm (cor. to 1 d.p.) Example 13 In △PQR, ∠R 90, PQ = 22 cm and QR =18 cm. Find ∠Q correct to the nearest 0.01. Solution ∵ ∴ QR cos Q PQ 18 cm 22 cm Q 35.10 (cor. to the nearest 0.01) Tangent Ratio Concept of Tangent Ratio The tangent ratio of an acute angle is defined as below: opposite side tan θ adjacent side θ adjacent side of opposite side of For a right-angled triangle with a given acute angle , the tangent ratio of is a constant. For example, 1 45 1 tan 45 = 3 2 45 2 1 2 3 = = 1 2 3 45 3 In ABC, C = 90, AC = 2.4 and BC = 3.2. A 2.4 B AC tan B BC 2.4 2.4 3.2 3.2 33 44 3.2 C AC is the adjacent side of A, and BC is the opposite side of A. Follow-up question 9 In the following figures, find tan θ. (a) (b) 5 5 θ θ 12 13 4 Solution (a) 12 tan θ 5 (b) 3 tan θ 4 3 Example 14 In △PQR, ∠R 90, PQ 37, PR 12 and RQ 35. Find the values of (a) tan∠P, (b) tan∠Q. (Give your answers in fractions.) Solution (a) QR tan P PR 35 12 PR (b) tan Q QR 12 35 Finding Tangent Ratio Using Calculators Find tan for a given angle In degree mode, use the key tan to find the value of tan . For example, the value of tan 45 can be obtained by keying: tan 45 EXE The answer is 1. Follow-up question 10 By using a calculator, find the values of the following expressions correct to 4 significant figures. tan 43 2 (a) 7 tan 51 (b) tan 57 Solution (a) (b) tan 51 = 1.234 89… 7 tan 51 8.644 (cor. to 4 sig. fig.) tan 43 2 2. (cor. to 4 sig. fig.) tan 43tan 257 1.904 tan 43 = 0.932 51…, tan 57 = 1.539 86… tan 57 1.904 (cor. to 4 sig. fig.) Example 15 By using a calculator, find the values of the following expressions correct to 4 significant figures. (a) tan 28.26 (b) tan 65.32 tan 46.15 Solution (a) tan 28.26 0.5375 (cor. to 4 sig. fig.) (b) tan 65.32 tan 46.15 2.265 (cor. to 4 sig. fig.) Find for a given value of tan In degree mode, use the keys SHIFT and tan to find the corresponding acute angle . For example, given that tan = 1, can be obtained by keying: SHIFT tan 1 EXE The answer is 45, i.e. = 45. Follow-up question 11 Find the acute angle in each of the following using a calculator. (Give your answers correct to 3 significant figures.) (a) tan θ 2.77 (b) tan θ 3 tan 20 Solution (a) 1. tan θ 2.77 1. tan θθ 70 2.77 (cor. to to the3 nearest 70.1 (cor. sig. fig.) degree) θ 20 (cor. the3 nearest 70.1 (cor. sig. fig.) degree) 2. tan θ 70 3 tan to to tan to 204=sig. 0.363fig.) 97… (b) tan 20 2. tan θθ 3 47.52 (cor. θ 47.52 (cor. 4 sig. fig.) 47.5 (cor. to 3tosig. fig.) Example 16 Find the acute angles in the following using a calculator. (a) (b) tan = 6.54, correct to the nearest degree. tan 2 tan 62 + 1, correct to 2 decimal places. (c) 9 tan 2, correct to 4 significant figures. Solution (a) tan 6.54 81 (cor. to the nearest degree) (b) tan 2 tan 62 1 78.14 (cor. to 2 d.p.) (c) 9 tan 2 12.53 (cor. to 4 sig. fig.) Using Tangent Ratio to Find Unknowns in Right-Angled Triangles We can use the tangent ratio to solve problems involving right-angled triangles. In ABC, B = 50, C = 90 and BC = 12 m. Find AC correct to 2 decimal places. AC AC AC tan B tan B tan B BC BC BC AC AC AC tan 50 tan 50 tan 50 12 m 12 12m m AC 12 tan 50 m AC AC 12 12 tan tan 50 50m m 14.30 m (cor. to 22 d.p.) (cor. d.p.) 14.30 14.30 m m (cor. to to 2 d.p.) A B 50 12 m C In PQR, R = 90, PR = 16 m and QR = 13 m. Find Q correct to 2 decimal places. AC tan B AC PR tan BC tan Q B QR BC AC tan 50 16 ACm tan 12 m tan 50 Q 12 m AC 13 12 m tan 50 m AC tan 50 m AC 12 12 tan 50 m to 2 d.p.) 50.91 (cor. (cor. 14.30 14.30 m m (cor. to to 2 2 d.p.) d.p.) P Q 16 m 13 m R Follow-up question 12 A In ABC, C 90, BC 2.4 cm and AC 7 cm. AC = 7 cm. Find B correct to 2 decimal places. 7 cm C 2.4 cm AC Solution tan B BC AC tan B 7 cm tan 51 BC BC 77 cm cm tan 51 B 7 BCcm cm BC 2.4 tan751 BC B 71.08 cm (cor. to 2 d.p.) tan 51 5.67 cm (cor. to 2 d.p.) B Example 17 In △PQR, ∠P = 65.2, ∠Q = 90 and PQ = 4 cm. Find the length of QR correct to 3 significant figures. Example 18 In △PQR, ∠P = 90, ∠R = 42.6 and PQ = 6.5 cm. Find the length of PR correct to 3 significant figures. Example 19 In △PQR, ∠Q = 90, PQ = 15 cm and QR = 12 cm. Find ∠P and ∠R correct to the nearest 0.1. Example 17 In △PQR, ∠P = 65.2, ∠Q = 90 and PQ = 4 cm. Find the length of QR correct to 3 significant figures. Solution ∵ ∴ QR tan P PQ QR tan 65.2 4 cm QR 4 tan 65.2 cm 8.66 cm (cor. to 3 sig. fig.) Example 18 In △PQR, ∠P = 90, ∠R = 42.6 and PQ = 6.5 cm. Find the length of PR correct to 3 significant figures. Solution ∵ ∴ PQ tan R PR 6.5 cm tan 42.6 PR 6.5 PR cm tan 42.6 7.07 cm (cor. to 3 sig. fig.) Example 19 In △PQR, ∠Q = 90, PQ = 15 cm and QR = 12 cm. Find ∠P and ∠R correct to the nearest 0.1. Solution QR PQ 12 cm 15 cm ∵ tan P ∴ P 38.7 (cor. to the nearest 0.1) ∵ tan R ∴ R 51.3 (cor. to the nearest 0.1) PQ QR 15 cm 12 cm Simple Applications of Trigonometric Ratios Solving Problems Involving Plane Figures For plane figures involving right-angled triangles, we can use sine, cosine or tangent ratio to find the length of an unknown side or the size of an unknown angle. Can you find the length of BC in the figure? Give your answer correct to 4 significant figures. A 8 cm B 40 35 D C In ABD ABD, In ,, In ABD AD AD AD sin35 sin 35 sin35 AB AB AB AD sin35 AD AB AB sin35 sin 35 88 8sin35 sin cm sin35 cm sin 35 35cm cm BD BD BD cos cos35 35 cos35 AB AB AB BD AB ABcos35 cos BD BD AB cos35 cos 35 35 88 8cos35 cos cm ......(1) (1) cos35 cm ...... (1) cos 35 35cm cm...... In ACD ACD,,, In In ACD AD AD AD tan40 tan 40 tan40 CD CD CD AD AD AD CD CD CD tan 40 AB AB BD BDAB ABcos35 cos35 88cos35 cos35cm cm...... ...... (1) (1) A InInIn ACD, ACD,, ACD 8 cm AD AD AD tan 40 tan40 tan40 35 B CD CD CD D AD AD CD CD tan 40 tan40 sin35 88sin 35 cm ...... (2) cm tan40 tan 40 40 C BC BCBD BDDC DC sin35 35 88sin cos35 35 cm 88cos cm tan40 40 tan 12.02cm cm (cor. 12.02 (cor.toto44sig. sig.fig.) fig.) Follow-up question 13 A In the figure, AD is the height , In ofABDABC , 5 cm 60 AB 5 cm, BAD 60 and ACB 45 . AD cos 60 45 AB B Find AC. D AB cos 60figures.) (Give your answer correct AD to 3significant 5 cos 60 cm Solution In ACD, 2.5 cm In , In ABD ABD, AD sin 45 AD AC cos 60 AB AD AC AD AB cos 60 sin 45 5 cos 60 cm 2.5 cm 2.5 cm sin 45 AD 3.54 cm (cor. (cor. tosig. 3 sig. fig.) to 3 fig.) sin 45 AC C Example 20 In the figure, BD is the height of △ABC, BC = 8 m, ∠BAC = 70 and ∠CBD = 60. Find AC. (Give your answer correct to 2 decimal places.) Solution In △BCD, BD cos 60 BC BD BC cos 60 8 cos 60 m 4m DC sin 60 BC DC BC sin 60 8 sin 60 m (1) In △ABD, BD tan 70 AD BD AD tan 70 4 m (2) tan 70 ∴ AC AD DC 4 8 sin 60 m tan 70 8.38 m (cor. to 2 d.p.) from (1) and (2) Example 21 In the figure, PQRS is a trapezium with ∠R = ∠S = 90, ∠Q = 52, PQ = 6 cm and PS = 5 cm. Find the area of PQRS correct to 2 decimal places. Solution Draw PT QR as shown in the figure. In △PQT, PT PQ PT PQ sin 52 6 sin 52 cm sin 52 QT cos 52 PQ QT PQ cos 52 6 cos 52 cm ∴ 1 Area of PQRS ( PS QR ) PT 2 1 (5 6 cos 52 5) 6 sin 52 cm 2 2 32.37 cm 2 (cor. to 2 d.p.) Example 22 The figure as shown is formed by a right-angled triangle PQR and two hemispheres with diameters PQ and RQ. ∠PQR = 90, ∠PRQ = 25 and PR = 16 cm. Find the perimeter of the figure correct to the nearest cm. Solution In △PQR, PQ sin 25 PR ∴ PQ PR sin 25 16 sin 25 cm QR cos 25 PR ∴ QR PR cos 25 16 cos 25 cm 1 1 Perimeter of the figure PQ QR 16 cm 2 2 1 1 16 sin 25 16 cos 25 16 cm 2 2 49 cm (cor. to the nearest cm) Solving Real-life Problems We can also use trigonometric ratios to solve real-life problems involving right-angled triangles. Let’s study the example on the next page. A rectangular advertising board is fixed to a vertical wall and is supported by two straight cable wires AB and AC, as shown in the figure. It is known that ABD = 30, B ACD = 60 and CD = 2 m. Find AC and AB. (Give your answers correct to 3 significant figures if necessary.) A 60 30 C D 2m A In ACD, In In ACD ACD,, In ACD, CD CD CD cos 60 cos 60 cos 60 CD AC AC AC cos 60 AC CD CD CD AC AC AC cos CD cos 60 AC cos 60 60 cos2260 2 m m m 2 cos 60 cos 60 cos 60 m cos 60 44 m 4m m 4m AD AD AD tan 60 tan 60 tan 60 AD CD CD CD tan 60 CD CD tan 60 AD AD AD CD CD tan 60 AD 2CD tan 60 tan 60 m 2 tan 60 m 2 tan 60 m 2 tan 60 m B 30 60 C 2m D CD CD CD cos cos 60 60 cos 60 AC AC AC CD CD CD AC AC AC cos 60 cos cos 60 60 22 2 m cos 60 m m cos 60 cos 60 B m 44 4m m AD AD AD tan tan 60 60 tan 60 CD CD CD AD CD tan 60 AD AD CD CD tan tan 60 60 tan 60 m 22 2 tan tan 60 60m m In ABD, AD sin 30 AB AD AB sin 30 A In ABD, 60 30AD Csin2 30 m D AB AD AB sin 30 2 tan 60 m sin 30 6.93 m (cor. to 3 sig. fig.) Follow-up question 14 A rectangular advertising board is fixed to a vertical wall and is supported by two straight cable wires AB and AC, as shown in the figure. It is known that B ACB 50, AC 1m and BC 1.7 m. Find ABC. (Give your answer correct to 3 significant figures.) A 1m 50 D 1.7 m C Follow-up question 14 (cont’d) In In ADC A,, ADC AD AD sin sin50 50 1 m AC AC C AD 50 AD AC AC sin sin50 50 B D 50 m 11sin sin 50 m 1.7 m Solution sin sin50 50m m DC DC In ADC, In ADC, cos cos 50 50 AC AC AD AD sin sin50 50 DC DC AC AC cos cos 50 50 AC AC 11cos cos 50 50m m AD AD AC AC sin sin50 50 cos 50 50m m cos 11sin sin50 50 m m sin sin50 50m m DC DC cos cos50 50 AC AC BD BC DC (1.7 cos 50) m Follow-up question 14 (cont’d) A 1m B Solution In ABD,, In ABD In ABD, 50 D 1.7 m C AD AD AD tan ABD ABD tan ABD tan BD BD BD sin50 50 sin 50 sin 1.7 cos 50 1.7 cos cos 50 50 1.7 ABD ABD 35.9 35.9 (cor.(cor. (cor. to sig. fig.) to 3to sig. fig.)fig.) ABD 35.9 33 sig. i.e. ABC ABC 35.9 35.9 i.e. ABC 35.9 i.e. Example 23 There is a fish pond between A and B. A man wants to go from A to B. He walks for 60 m from A to C, then turns 75 clockwisely and walks for 42 m from C to B. If ∠BAC = 30, find AB. (Give your answer correct to 1 decimal place.) Solution Draw CD AB as shown in the figure. In △ACD, AD cos 30 AC ∴ AD AC cos 30 60 sin 30 m (1) ACD 180 90 30 ( sum of △) 60 BCD 180 75 60 (adj. s on st. line) 45 In △BCD, DB BC DB BC sin BCD 42 sin 45 m (2) sin BCD ∴ ∴ AB AD DB (60 cos 30 42 sin 45) m 81.7 m (cor. to 1 d.p.) Example 24 Find the values of the following trigonometric ratios using the quarter of the unit circle as shown. (Give your answers correct to 1 decimal place.) (a) (c) sin 65 cos 46 (b) sin 12 (d) cos 84 Solution (a) Construct line segment OP such that OP makes an angle 65 with the positive x-axis. sin 65 y -coordinate of P 0.9 (b) Construct line segment OQ such that OQ makes an angle 37 with the positive x-axis. sin 12 y -coordinate of Q 0.2 (c) Construct line segment OR such that OR makes an angle 46 with the positive x-axis. cos 46 x-coordinate of R 0.7 (d) Construct line segment OS such that OS makes an angle 84 with the positive x-axis. cos 84 x-coordinate of S 0.1 Trigonometric Ratios of Special Angles In general, the the values shown In fact, the exact values of the Can you find out value of the calculator trigonometric ratios screen of someare sinon 60°? approximations only. special angles such as 30°, 45° and 60° can be deduced from the properties of triangles. With a calculator, I can evaluate sin 60° = 0.866 025 403... First, let’s review on the trigonometric ratios and the Pythagoras’ theorem. Consider right-angled triangle ABC, we have 1. sin B b c c a cos B B c a b tan B a 2. By Pythagoras’ theorem, c 2 a2 b2 A b C Using the above knowledge and considering the following triangles, we can find the exact values of the trigonometric ratios of 30°, 45° and 60°. R C 60° 45° A 45° 1 B 2 2 1 P 60° 60° 2 Q Trigonometric Ratios of 45° C First, find the exact values Consider isosceles right-angled can apply of Sincelet’s = 90 Bthe °, we the trigonometric ratios 45AC. °. triangle ABC theorem on the right. Pythagoras’ toof find 45° 1 A 45° 1 2 2 AC (AB (Pyth. theorem) 1 )2 ______ 1 ) (BC We have 1 2 BC or sin 45° = 2 2 AC cos 45° = AB 1 2 or 2 AC 2 BC tan 45° = 1 AB B Trigonometric Ratios of 60° and 30° First construct asides, perpendicular line Now, Since we PRS can and find QRS the trigonometric aretrigonometric two congruent Now, let’s try to find the We have found the exact values of from R PS = QS (corr. s). Consider the triangle PQR. First, find and PRS. PQR isfind anPS equilateral triangle. Then, RS. and meet PQ at S. ratios right-angled of 60 and triangles, 30 . ratios of special angles and 30°. ° ° of60° the trigonometric ratios 45°. Consider △PRS. PS ___ 30° 1 and PRS ____ 2 2 RS (RP 3 (Pyth. theorem) 1 ) ____ 2 ) ( PS R We have 60° P 60° 60° S 2 2 1 sin 30 PS PS 1 cos 60 RP 2 RS 3 cos 30 PR 2 RS tan 60 3 PS 1 3 PS or tan 30 RS 3 3 RP 2 2 2 3 sin 60 RS Q PR 2 The table below summarizes the trigonometric ratios of the special angles 30°, 45° and 60°. Trigonometric ratio θ 30 45 60 sin θ 1 2 1 2 or 2 2 3 2 cos θ 3 2 1 2 or 2 2 1 2 tan θ 1 3 or 3 3 1 They are useful when we need to find the values of trigonometric expressions involving special angles. 3 Without using a calculator, find the value of the expression sin 30° tan 60° + sin 60°. 1 33 sin 30 tan 60 sin 60 33 2 22 3 3 2 2 3 Follow-up question 15 Find the values of the following expressions without using a calculator. sin 45cos 45 1 2 (a) (b) tan 30 cos 60 tan 2 60 Solution 1 1 sin 45cos 45 2 2 (a) 1 cos 60 2 1 2 1 2 1 Follow-up question 15 (cont’d) Find the values of the following expressions without using a calculator. sin 45cos 45 1 2 (a) (b) tan 30 cos 60 tan 2 60 Solution (b) 2 1 1 tan 30 2 tan 60 3 2 1 1 3 3 2 3 1 3 2 Since exact valuesangles of the in we canthe find the acute trigonometric ratios ofequations special simple trigonometric angles are known, without using a calculator. For example: (a) 2cos 1 1 cos 2 60 (b) 2 tan 2 sin 45 2 tan 2 1 cos 60° = __ 2 2 2 2 tan 2 tan 1 45 tan 45° = 1 Follow-up question 16 Find the acute angles in each of the following equations without using a calculator. 1 (a) 2sin cos 45 (b) tan sin 60 0 2 Solution (a) 2sin cos 45 2 2 1 sin 2 2 sin 30 Follow-up question 16 (cont’d) Find the acute angles in each of the following equations without using a calculator. 1 (a) 2sin cos 45 (b) tan sin 60 0 2 Solution 1 (b) tan sin 60 0 2 1 3 tan 0 2 2 1 3 tan 2 2 tan 3 60 Example 25 Find the values of the following expressions without using a calculator. (a) cos 60 tan 30 tan 60 (b) tan 45 4 sin 30 cos 2 30 (c) tan 60 sin 60 sin2 45 1 4 (b) Solution (a) 1 1 3 2 3 1 1 2 (c) 1 2 cos 60 tan 30 tan 60 1 2 tan 45 4 sin 30 2 cos 2 30 3 2 1 2 3 4 4 3 1 tan 60 sin 60 sin 2 45 3 2 2 3 1 2 2 1 2 Example 26 Find the acute angle in each of the following equations without using a calculator. (a) (b) cos cos2 45 3 tan ( 10) 1 Solution (a) cos cos 2 45 1 cos 2 1 cos 2 60 2 (b) 3 tan ( 10) 1 1 3 10 30 40 tan ( 10) Example 27 Referring to the figure, find the lengths of the following line segments without using a calculator. (Leave your answers in surd form.) (a) AC (b) DC Solution (a) Consider △ABC. AC sin 45 BC AC BC sin 45 2 10 cm 2 5 2 cm (b) Consider △ACD. AC DC AC DC tan 60 5 2 cm 3 tan 60 Finding Trigonometric Ratios by Constructing Right-Angled Triangles 4 If sin = 5, how can I find cos and tan ? Do I need to evaluate first? You can find cos and tan by the following steps without evaluating . Step 1 Construct a right-angled triangle ABC with A =θand B = 90°. Step 2 5 opposite side of θ 4 Since sinθ= , we set BC = 4 and AC = 5. 5 5 hypotenuse Step 3 Find the unknown side AB by Pythagoras’ theorem. AB AC 2 BC 2 52 42 3 4 3 Step 4 Find the other two trigonometric ratios by their definitions. AB cos AC 3 5 BC tan AB 4 3 In general, if one of the trigonometric ratios of an acute angle θ is given, we can follow these steps to find the other two trigonometric ratios without evaluating θ. 5 4 3 Follow-up question 17 It is given that tanθ= 0.5, whereθis an acute angle. Find the values of sinθand cosθwithout evaluatingθ. (Give your answers in surd form.) Solution 5 1 tan 0.5 10 2 1 Construct △ABC as shown with tanθ= . 2 By Pythagoras’ theorem, BC AC AB 2 12 2 2 5 2 C 5 1 A 2 B Follow-up question 17 (cont’d) It is given that tanθ= 0.5, whereθis an acute angle. Find the values of sinθand cosθwithout evaluatingθ. (Give your answers in surd form.) Solution AC By definition, sin BC 1 5 or 5 5 2 5 5 1 AB cos BC C A 2 5 or 5 2 B Example 28 2 , where is an acute angle. Find the values of 5 cos and tan without evaluating . (Leave your answers in surd form.) It is given that sin Solution 2 Construct △ABC as shown with sin . 5 By Pythagoras’ theorem, AB AC 2 BC 2 52 2 2 21 By definition, AB cos AC 21 5 tan BC AB 2 21 2 21 or 21 Example 29 It is given that cos 0.25, where is an acute angle. Find the values of sin and tan without evaluating . (Leave your answers in surd form.) Solution cos 0.25 25 100 1 4 1 Construct △ABC as shown with cos . 4 By Pythagoras’ theorem, BC AC 2 AB2 42 12 15 By definition, BC sin AC 15 4 BC AB 15 1 15 tan Example 30 40 It is given that tan , where is an acute angle. Find the value of 9 sin + cos without evaluating . (Give your answer in fraction.) Solution 40 Construct △ABC as shown with tan . 9 By Pythagoras’ theorem, AC AB 2 BC 2 92 402 1681 41 By definition, sin cos ∴ BC AC 40 41 AB AC 9 41 40 9 sin cos 41 41 49 41 Trigonometric Identities Basic Trigonometric Identities Complete the table. What can you find? sin cos θ sin θ cos θ tan θ 30° 1 2 3 2 1 3 1 3 1 45° 1 2 1 2 1 1 1 60° 3 2 1 2 3 3 1 sin 2 cos 2 I find that 2 3045 cos 2 30 1 I find that sinsin 60 30 45 tan 60 2 2 30 sincos 4560 cos 45 1 45 30 sin 2 60 cos 2 60 1 sin Is tan always true? cos Is sin 2 cos 2 1 always true? Let’s study the following proof. Consider the right-angled triangle ABC as shown. b a b sin , cos , tan c c a Then (i) b sin c a cos c b a tan Consider the right-angled triangle ABC as shown. b a b sin , cos , tan c c a Then 2 (ii) b a sin 2 cos 2 c c 2 b2 a2 c2 c2 2 c 1 c2 = b2 a2 (Pyth. theorem) We have the following two basic trigonometric identities. sin θ tan θ cos θ sin2θ cos2θ 1 Note that sin2θ cos 2θ 1 can also be written as sin2θ 1 cos 2θ or cos 2θ 1 sin2θ. Simplify the following expressions. (b) 1 tan 2 (a) tan sin (a) tan tan 1 sin sin sin 1 cos sin 1 cos sin tan = _____ cos Simplify the following expressions. (a) tan (b) 1 tan 2 sin 2 sin 2 (b) 1 tan 1 cos 2 sin tan2 = (tan )2 and tan = _____ cos cos 2 sin2 cos 2 12 cos cos2 sin2 = 1 Follow-up question 18 Simplify the following expressions. (a) (1 sin 2 ) tan 2 (b) cos 2 1 1 sin 2 (b) cos 2 1 1 sin 2 Solution (a) (1 sin 2 ) tan 2 cos 2 tan 2 2 sin 2 cos cos 2 sin2 (1 cos 2 ) 1 sin 2 sin 2 cos 2 tan 2 Example 31 Simplify the following expressions. (a) sin tan cos 2 (b) 4 sin 2 4 cos 2 (c) 6 3 sin 2 4 cos 2 4 Solution (a) (b) sin sin 2 tan cos sin cos 2 cos sin sin cos 1 cos 4 sin 2 4 cos 2 4(sin 2 cos 2 ) 4(1) 4 (c) 6 3 sin 2 6 3(1 cos 2 ) 2 2 sin cos 1) (∵ 2 2 4 cos 4 4 cos 4 3 3 cos 2 4 cos 2 4 3(cos 2 1) 4(cos 2 1) 3 4 Example 32 Simplify the following expressions. sin 2 (a) 1 tan 2 sin cos (b) 1 sin 2 Solution (a) sin 2 1 2 1 1 sin 2 tan sin 2 cos 2 2 cos 2 1 sin sin 2 1 cos 2 sin 2 (b) (∵ sin 2 cos 2 1 ) sin cos sin cos 2 1 sin cos 2 sin cos tan Example 33 8 It is given that sin , where is an acute angle. 17 (a) By using the trigonometric identities, find the values of cos and tan . (b) Hence, find the value of 5 sin 2 cos . tan (Give your answers in fractions.) Solution (a) ∵ cos 2 1 sin 2 ∴ cos 1 sin 2 8 1 17 1 64 289 225 289 15 17 2 sin tan cos 8 17 15 17 8 15 (b) 5 sin 2 cos tan 8 15 5 2 17 17 8 15 40 30 17 17 8 15 10 17 8 15 75 68 Trigonometric Ratios of Complementary Angles Consider the right-angled triangle ABC as shown. a sin θ = sin b c cos θ = cos b a tan θ = tan c c sin sin (90° (90° θ ) = b a cos cos (90° (90° θ ) = b c tan tan (90° (90° θ ) = a = = = 1 Using the trigonometric identities, find the acute angle in each of the following. 1 (a) sin θ cos 35 (b) tan ( 90 θ ) tan 27 (a) sin cos 35 sin (90 35) sin 55 55 cos = sin (90° ) Using the trigonometric identities, find the acute angle in each of the following. 1 (a) sin θ cos 35 (b) tan ( 90 θ ) tan 27 1 (b) tan ( 90 θ ) tan 27 1 1 1 tan (90° ) = _____ tan tan tan 27 tan tan 27 27 Using the trigonometric identities, find the acute angle in each of the following. 1 (a) sin θ cos 35 (b) tan ( 90 θ ) tan 27 (b) Alternative Solution tan ( 90 θ ) 1 tan 27 tan (90 θ ) tan (90 27) 90 θ 90 27 27 1 tan (90° ) = _____ tan Follow-up question 19 Using the trigonometric identities, find the acute angle in each of the following. (a) cos (90 ) sin 42 (b) tan tan 35 tan 45 Solution (a) cos (90 ) sin 42 (b) tan tan 35 tan 45 sin sin 42 tan tan 35 1 1 tan tan 35 42 tan (90 35) tan 55 55 Example 34 24 It is given that tan , where is an acute angle. By using the 7 trigonometric identities, find the values of sin and cos . (Give your answers in fractions.) Solution ∵ ∴ 24 tan 7 sin 24 cos 7 7 sin 24 cos (*) 49 sin 2 576 cos 2 49(1 cos 2 ) 576 cos 2 625 cos 2 49 49 2 cos 625 7 cos 25 (∵ sin 2 cos 2 1 ) ∵ ∴ 7 sin 24 cos 7 7 sin 24 25 24 sin 25 (from (*) ) Example 35 It is given that cos 1 , where is an acute angle. Using the 3 trigonometric identities, find the value of 3 cos2 sin2 . (Give your answer in fraction.) Solution 3 cos 2 sin 2 3 cos 2 (1 cos 2 ) 4 cos 2 1 2 1 4 1 3 4 1 9 5 9 (∵ sin 2 cos 2 1 ) Proofs of Simple Trigonometric Identities We have learnt fivetotrigonometric can use them prove other identities. trigonometric identities. sin cos (ii) sin2 cos 2 1 (i) tan (iii) sin(90 ) cos (iv) cos( 90 ) sin (v) tan( 90 ) 1 tan Prove that 1 cos sin . sin tan 1 L.H.S. sin sin R.H.S. 1 sin 2 sin cos 2 sin cos cos tan ∵ L.H.S. R.H.S. 1 sin2 = cos2 cos sin 1 cos tan cos tan 1 cos ______ _____ = sin sin _____ cos 1 = _____ tan 1 cos sin sin tan Follow-up question 20 1 Prove that 1 2 tan 2 θ. sin (90 θ ) Solution L.H.S. 1 1 1 sin 2 ( 90 θ ) R.H.S. tan 2 1 cos 2 L.H.S. R.H.S. (1 cos 2 ) cos 2 sin 2 cos 2 tan 2 1 1 2 tan 2 sin (90 ) Example 36 Find the acute angle in each of the following equations by using the trigonometric identities. (a) 1 tan tan 66 (b) cos ( 30) sin 2 Solution tan (a) tan tan ∴ 1 tan 66 tan (90 66) tan 24 24 (b) cos ( 30) sin 2 cos ( 30) cos 90 2 ∴ 30 90 3 60 2 40 2 Example 37 Simplify the following expressions. (a) sin (90 ) cos (90 ) tan (90 ) (b) cos (90 ) sin sin 2 (90 ) Solution (a) (b) 1 sin (90 ) cos (90 ) tan (90 ) cos sin tan 1 cos sin sin cos cos cos sin sin cos 2 cos (90 ) sin sin 2 (90 ) sin sin cos 2 sin 2 cos 2 1 Example 38 Find the values of the following expressions. 1 sin 2 37 (a) sin 2 53 (b) cos 21 tan 69 sin 21 Solution (a) 1 sin 2 37 cos 2 37 2 sin 53 sin 2 53 sin 2 (90 37) sin 2 53 sin 2 53 sin 2 53 1 1 sin 21 (b) cos 21 tan 69 sin 21 cos 21 tan (90 69) 1 cos 21 sin 21 tan 21 1 cos 21 sin 21 sin 21 cos 21 cos 21 cos 21 sin 21 sin 21 cos 21 cos 21 0 Example 39 Prove the following trigonometric identities. sin cos 1 (a) cos sin sin sin (90 ) (b) sin tan 1 cos cos Solution (a) sin cos L.H.S. cos sin sin 2 cos 2 sin cos 1 sin cos 1 sin sin (90 ) 1 sin cos R.H.S. ∵ ∴ L.H.S. R.H.S. sin cos 1 cos sin sin sin (90 ) (b) 1 cos cos 1 cos 2 cos sin 2 cos sin sin cos sin tan L.H.S. sin tan R.H.S. ∵ L.H.S. R.H.S. ∴ sin tan 1 cos cos Example 40 Prove that (sin cos )2 1 2 sin cos . Solution L.H.S. (sin cos ) 2 sin 2 2 sin cos cos 2 (sin 2 cos 2 ) 2 sin cos 1 2 sin cos R.H.S. 1 2 sin cos ∵ L.H.S. R.H.S. ∴ (sin cos ) 2 1 2 sin cos Extra Teaching Examples Example 7 (Extra) In △ABC, ∠ADC = 90, AB = 7.5 cm, AC = 6.5 cm and AD = 5.5 cm. Find ∠BAC, ∠B and ∠C correct to 1 decimal place. Solution ∵ AD sin B AB 5.5 cm 7.5 cm ∴ B 47.2 (cor. to 1 d.p.) ∵ AD sin C AC 5.5 cm 6.5 cm ∴ C 57.8 (cor. to 1 d.p.) In △ABC, ( sum of △) BAC 180 B C 180 47.2 57.8 75.0 (cor. to 1 d.p.) Example 13 (Extra) In △ABD, ∠D 90, AB 11 cm, AC 7.8 cm and AD 6 cm. Find ∠BAC correct to the nearest 0.1. Solution ∵ ∴ AD cos DAC AC 6 cm 7.8 cm DAC 39.7 ∵ AD cos DAB AB 6 cm 11 cm ∴ DAB 56.9 ∴ BAC DAB DAC 56.9 39.7 17.2 (cor. to the nearest 0.1) Example 19 (Extra) In △ABC, ∠ABD = ∠DBC = 25, ∠C = 90 and BC = 3.5 m. Find AD correct to 2 decimal places. Solution In △ABC, ∵ ∴ AC tan ABC BC AC tan (25 25) 3.5 m AC 3.5 tan 50 m In △BCD, ∵ ∴ ∴ DC tan DBC BC DC tan 25 3.5 m DC 3.5 tan 25 m AD AC DC (3.5 tan 50 3.5 tan 25) m 2.54 m (cor. to 2 d.p.) Example 21 (Extra) The figure shows a quadrilateral ABCD with ∠A = 115, ∠B = 90, ∠C = 80, AB =13 cm and AD = 18 cm. Find BC correct to the nearest cm. Solution Draw DF BC and AE DF as shown in the figure. In △ADE, cos DAE AE AD ∴ AE AD cos DAE 18 cos (115 90) cm 18 cos 25 cm ∴ BF AE 18 cos 25 cm ……(1) DE sin DAE AD ∴ DE AD sin DAE 18 sin (115 90) cm 18 sin 25 cm DF DE EF DE AB (18 sin 25 13) cm In △CDF, DF tan 80 CF ∴ DF CF tan 80 18 sin 25 13 cm tan 80 ……(2) ∴ BC BF CF 18 sin 25 13 18 cos 25 cm tan 80 20 cm (cor. to the nearest cm) Example 22 (Extra) In the figure, sector OPQ is inscribed in rectangle ABCD. Given that AB =10 cm and BC = 14 cm, find the area of sector OPQ correct to 1 decimal place. Solution In △OBQ, OB cos QOB OQ 10 cm 2 14 cm ∴ QOB 69.1 QOB POA POQ QOB POA 180 (adj. s on st. line) POQ 2QOB 180 POQ 2(69.1) 180 POQ 41.8 ∴ 41.8 14 2 cm 2 Area of sector OPQ 360 71.6 cm 2 (cor. to 1 d.p.) Example 27 (Extra) The figure shows two shadows AD and BD of a tree CD at 9:00 a.m. and 4:30 p.m. respectively. If the height of the tree is 8 m, find the distance between A and B. (Leave your answer in surd form.) Solution Consider △ACD. CD tan 45 AD CD AD tan 45 8m 1 8m Consider △BCD. CD tan 30 BD CD BD tan 30 8m 1 3 8 3 m Distance between A and B AD DB (8 8 3 ) m 8(1 3 ) m Example 38 (Extra) Find the value of 3 3 tan 2 26 sin 2 64 3 cos 2 26 . Solution 3 3 tan 2 26 sin 2 64 3 cos 2 26 3 3 tan 2 26 cos 2 (90 64) 3 cos 2 26 sin 2 26 2 2 3 3 cos 26 3 cos 26 2 cos 26 3 3 sin 2 26 3 cos 2 26 3 3(sin 2 26 cos 2 26) 3 3(1) 0 Example 39 (Extra) Prove that 1 cos2 (90 ) sin 2 tan 2 (90 ) . Solution L.H.S. 1 cos 2 (90 ) 1 sin 2 cos 2 ∵ ∴ R.H.S. sin 2 tan 2 (90 ) 1 tan 2 1 2 sin sin 2 cos 2 2 cos 2 sin sin 2 cos 2 sin 2 L.H.S. R.H.S. 1 cos2 (90 ) sin 2 tan 2 (90 )