Practice Set 8.1 Practice Set 8.2 Problem Set 8
Practice Set 8.1Question 1. In the Fig. 8.12, ∠R is the right angle of ΔPQR. Write the following ratios.(i) sin P (ii) cos Q(iii) tan P (iv) tan Q Answer: For any right-angled triangle, sinθ = Opposite side Side/Hypotenuse cosθ = Adjacent sideSide/Hypotenuse tanθ = sinθ/cosθ = Opposite side Side/Adjacent sideSide cotθ = 1/tanθ = Adjacent sideSide/Opposite side Side secθ = 1/cosθ = Hypotenuse/Adjacent sideSide cosecθ = 1/sinθ = Hypotenuse/Opposite side Side In the given triangle let us understand, the Opposite side and Adjacent sidesides. So for ∠ P, Opposite side Side = QR Adjacent sideSide = PR So, for ∠ Q, Opposite side Side = PR Adjacent sideSide = QR In general for the side Opposite side to the 90° angle is the hypotenuse. So, for Δ PQR, hypotenuse = PQ (i) sin P = Opposite side Side/Hypotenuse = QR/PQ (ii) cos Q = Adjacent sideSide/Hypotenuse = QR/PQ (iii) tan P = sinθ/cosθ = Opposite side Side/Adjacent sideSide = QR/PR (iv) tan Q = sinθ/cosθ = Opposite side Side/Adjacent sideSide = PR/QR Question 2. In the right angled ΔXYZ, ∠XYZ = 900 and a,b,c are the lengths of the sides as shown in the figure. Write the following ratios,(i) sin X (ii) tan Z(iii) cos X (iv) tan X. Answer: For any right-angled triangle, sinθ = Opposite side Side/Hypotenuse cosθ = Adjacent Side/Hypotenuse tanθ = sinθ/cosθ = Opposite Side/Adjacent Side In the given triangle let us understand, the Opposite side and Adjacent side So for ∠ X, Opposite Side = YZ = a Adjacent Side = XY = b So for ∠ Z, Opposite Side = XY = b Adjacent Side = YZ = a In general for the side Opposite side to the 90° angle is the hypotenuse. So for Δ XYZ, hypotenuse = XZ = c (i) sin X = Opposite side Side/Hypotenuse = YZ/XZ = a/c (ii) tan Z = sinθ/cosθ = Opposite Side/Adjacent Side = XY/YZ = b/a (iii) cos X= Adjacent Side/Hypotenuse = XY/XZ = b/c (iv) tan X = sinθ/cosθ = Opposite Side/Adjacent Side = YZ/XY = a/b Question 3. In right angled ΔLMN, ∠LMN =900, ∠L = 500 and ∠N = 400write the following ratios.(i) sin 50° (ii) cos 50°(iii) tan 40° (iv) cos 40° Answer: For any right-angled triangle, sinθ = Opposite side Side/Hypotenuse cosθ = Adjacent sideSide/Hypotenuse tanθ = sinθ/cosθ = Opposite side Side/Adjacent sideSide cotθ = 1/tanθ = Adjacent sideSide/Opposite side Side secθ = 1/cosθ = Hypotenuse/Adjacent sideSide cosecθ = 1/sinθ = Hypotenuse/Opposite side Side In the given triangle let us understand, the Opposite side and Adjacent sidesides. So for ∠ 50°, Opposite side Side = MN Adjacent sideSide = LM So for ∠ 40°, Opposite side Side = LM Adjacent sideSide = MN In general, for the side Opposite side to the 90° angle is the hypotenuse. So, for Δ LMN, hypotenuse = LN (i) sin 50° = Opposite side Side/Hypotenuse = MN/LN (ii) cos 50° = Adjacent sideSide/Hypotenuse = LM/LN (iii) tan 40° = sinθ/cosθ = Opposite side Side/Adjacent sideSide = LM/MN (iv) cos 40° = Adjacent sideSide/Hypotenuse = MN/LN Question 4. In the figure 8.15 ∠PQR = 900, ∠PQS = 900, ∠PRQ = α and ∠QPS = θ Write the following trigonometric ratios.i. sin α, cos α, tan αii. sin θ, cos θ, tan θ Answer: For any right-angled triangle, sinθ = Opposite side Side/Hypotenuse cosθ = Adjacent sideSide/Hypotenuse tanθ = sinθ/cosθ = Opposite side Side/Adjacent sideSide cotθ = 1/tanθ = Adjacent sideSide/Opposite side Side secθ = 1/cosθ = Hypotenuse/Adjacent sideSide cosecθ = 1/sinθ = Hypotenuse/Opposite side Side (i) In the given triangle let us understand, the Opposite side and Adjacent sidesides. So, for Δ PQR, So, for ∠ α, Opposite side Side = PQ Adjacent sideSide = QR In general for the side Opposite side to the 90° angle is the hypotenuse. So, for Δ PQR, hypotenuse = PR sin α = Opposite side Side/Hypotenuse = PQ/PR cos α = Adjacent sideSide/Hypotenuse = QR/PR tan α = sinθ/cosθ = Opposite side Side/Adjacent sideSide = PQ/QR (ii) In the given triangle let us understand, the Opposite side and Adjacent sidesides. So for Δ PQS, So for ∠θ, Opposite side Side = QS Adjacent sideSide = PQ In general for the side Opposite side to the 90° angle is the hypotenuse. So for Δ PQS, hypotenuse = PS sinθ = Opposite side Side/Hypotenuse = QS/PS cosθ = Adjacent sideSide/Hypotenuse = PQ/PS tanθ = sinθ/cosθ = Opposite side Side/Adjacent sideSide = QS/PQ Question 1. In the following table, a ratio is given in each column. Find the remaining two ratios in the column and complete the table. Answer: For first column: cosθ = 35/37 Adjacent side= 35, Hypotenuse = 37 By Pythagoras Theorem Hypotenuse2 = Opposite side2 + Adjacent2 Opposite side2 = Hypotenuse2 - Adjacent2 = 372 - 352 = 1369 – 1225 Opposite side2 = 144 Opposite side = 12 For second column: Opposite side = 11 Hypotenuse = 61 By Pythagoras Theorem Hypotenuse2 = Opposite side2 + Adjacent2 Adjacent2 = Hypotenuse2 - Opposite side2 = 612 - 112 = 3721 – 121 Adjacent2 = 3600 Adjacent side= 60 For third column: Opposite side = 1 Adjacent side= 1 By Pythagoras Theorem Hypotenuse2 = Opposite side2 + Adjacent2 = 1 + 1 Hypotenuse2 = 2 Hypotenuse = √2 For fourth column: Opposite side = 1 Hypotenuse = 2 By Pythagoras Theorem Hypotenuse2 = Opposite side2 + Adjacent2 Adjacent2 = Hypotenuse2 - Opposite side2 = 22 - 12 = 4 – 1 Adjacent2 = 3 Adjacent side= √3 For fifth column: Adjacent side= 1 Hypotenuse = √3 By Pythagoras Theorem Hypotenuse2 = Opposite side2 + Adjacent2 Opposite side2 = Hypotenuse2 - Adjacent2 = (√3)2 - 12 = 3 – 1 Opposite side2 = 2 Opposite side = √2 For sixth column: Opposite side = 21 Adjacent side= 20 By Pythagoras Theorem Hypotenuse2 = Opposite side2 + Adjacent2 = 212 + 202 Hypotenuse2 = 841 Hypotenuse = 29 For seventh column: Opposite side = 8 Adjacent side= 15 By Pythagoras Theorem Hypotenuse2 = Opposite side2 + Adjacent2 = 82 + 152 Hypotenuse2 = 289 Hypotenuse = 17 For eighth column: Opposite side = 3 Hypotenuse = 5 By Pythagoras Theorem Hypotenuse2 = Opposite side2 + Adjacent2 Adjacent2 = Hypotenuse2 - Opposite side2 = 52 - 32 = 25 – 9 Adjacent2 = 16 Adjacent side= 4 For ninth column: Opposite side = 1 Adjacent side= 2√2 By Pythagoras Theorem Hypotenuse2 = Opposite side2 + Adjacent2 = 12 + (2√2)2 Hypotenuse2 = 9 Hypotenuse = 3 Question 2. Find the values of – 5sin 300 + 3tan450 Answer: We know, sin 30° = 1/2 tan 45° = 1 ⟹ 5sin 30° + 3tan 45° ⟹ ⟹ 2.5 + 3 ⟹ 5.5 Question 3. Find the values of – Answer: We know, tan 60° = √3 sin 60° = √3/2 ⟹ ⟹ ⟹ ⟹ ⟹ = 93/20 Question 4. Find the values of – 2sin 300 + cos 00 + 3sin 900 Answer: We know, sin 30° = 1/2 cos 0° = 1 sin 90° = 1 ⟹ ⟹ ⟹ 1 + 1 + 1 = 3 Question 5. Find the values of – Answer: We know, tan 60° = √3 sin 60° = √3/2 cos 60° = 1/2 ⟹ ⟹ ⟹ Question 6. Find the values of – cos2450 + sin2300 Answer: We know, cos 45° = 1/√2 sin 30° = 1/2 ⟹ ⟹ ⟹ Question 7. Find the values of – cos 600× cos 300 + sin600 × sin300 Answer: We know, sin 30° = 1/2 sin 60° = √3/2 cos 60° = 1/2 cos 30° = √3/2 ⟹ ⟹ ⟹ ⟹ Question 8. If sinθ = 4/5 then find cosθ. Answer: We know, sinθ = Opposite side/Hypotenuse Given: sinθ = 4/5 Opposite side = 4 Hypotenuse = 5 By Pythagoras Theorem Hypotenuse2 = Opposite side2 + Adjacent2 Adjacent2 = Hypotenuse2 - Opposite side2 = 52 - 42 = 25 – 16 = 9 Adjacent2 = 9 Adjacent side= 3 cosθ = Adjacent side/Hypotenuse = 3/5 Question 9. If then find sinθ Answer: We know, cosθ = Adjacent side/Hypotenuse Adjacent side = 15 Hypotenuse = 17 By Pythagoras Theorem Hypotenuse2 = Opposite side2 + Adjacent2 Opposite side2 = Hypotenuse2 - Adjacent2 = 172 - 152 = 289 – 225 = 64 Opposite side2 = 64 Opposite side = 8 sinθ = Opposite side /Hypotenuse = 8/17 Question 1. Choose the correct alternative answer for following multiple choice questions.Which of the following statements is true? A. sin θ = cos(90-θ)B. cos θ = tan(90-θ)C. sin θ = tan(90-θ)D. tan θ = tan(90-θ) Answer: Let us consider the given triangle, In this Δ PMN, For ∠θ, Opposite side = PM Adjacent side= PN For ∠ (90 –θ) Opposite side = MN Adjacent side = PM sinθ = Opposite side/Hypotenuse = PM/PN …………………… (i) cos (90-θ) = Adjacent/Hypotenuse = PM/PN ……………………. (ii) RHS of equation (i) and (ii) are equal ∴ sinθ = cos (90-θ) So Option A is correct. Question 2. Choose the correct alternative answer for following multiple choice questions.Which of the following is the value of sin 90°? A. B. 0C. D. 1Answer: We know that the value of sin 90° = 1 So option D is correct. Question 3. Choose the correct alternative answer for following multiple choice questions. 2tan 450 + cos 450 – sin 450 =? A. 0B. 1C. 2D. 3Answer: We know that, tan 45° = 1 We also know that cos 45° = sin 45° So, ⟹ 2 × 1 + cos 45° - cos 45° = 2 So the correct option is C. Question 4. Choose the correct alternative answer for following multiple choice questions. A. 2B. -1C. 0D. 1Answer: We know the identity that, sinθ = cos (90 –θ) sin 62° = cos (90 – 62) = cos 28° Therefore [cos 28°/cos 28°] = 1 So option D is correct. Question 5. In right angled ΔTSU, TS = 5, ∠S = 900, SU =12 then find sin T, cos T, tan T. Similarly find sin U, cos U, tan U. Answer: By applying Pythagoras theorem to given triangle we have, sinT cosT tanTSimilarly, Question 6. In right angled ΔYXZ, ∠X = 900, XZ = 8cm, YZ =17cm, find sin Y, cos Y, tan Y, sin Z, cos Z, tan Z. Answer: For any right-angled triangle, sinθ = Opposite side /Hypotenuse cosθ = Adjacent side/Hypotenuse tanθ = sinθ/cosθ = Opposite side/Adjacent side cotθ = 1/tanθ = Adjacent side/Opposite side secθ = 1/cosθ = Hypotenuse/Adjacent side cosecθ = 1/sinθ = Hypotenuse/Opposite side In the given triangle let us understand, the Opposite side and Adjacent sides. So for ∠ Y, Opposite side = XZ =8 Adjacent side= XY So for ∠ Z, Opposite side = XY Adjacent side = XZ = 8 In general for the side Opposite side to the 90° angle is the hypotenuse. So for Δ TSU, By Pythagoras Theorem YZ2 = XZ2 + XY2 XY2 = 172 - 82 = 289 - 64 = 225 XY = 15 (i) sin Y = Opposite side/Hypotenuse = XZ/YZ = 8/17 (ii) cos Y = Adjacent side/Hypotenuse = XY/YZ = 15/17 (iii) tan Y = sinθ/cosθ = Opposite side/Adjacent side = XZ/XY = 8/15 (i) sin Z = Opposite side/Hypotenuse = XY/YZ = 15/17 (ii) cos Z = Adjacent side/Hypotenuse = XZ/YZ = 8/17 (iii) tan Z = sinθ/cosθ = Opposite side/Adjacent side = XZ/XY = 8/15 Question 7. In right angled ΔLMN, if ∠N = θ, ∠M = 900, cosθ = 24/25 find sinθ and tanθ Similarly, find (sin2θ) and (cos2θ). Answer: Give: cosθ = 24/25 cosθ = Adjacent side/Hypotenuse Adjacent side = 24 Hypotenuse = 25 By Pythagoras Theorem Hypotenuse2 = Opposite side2 + Adjacent2 Opposite side2 = Hypotenuse2 - Adjacent2 = 252 - 242 = 625 – 576 = 49 Opposite side2 = 49 Opposite side = 7 sinθ = Opposite side/Hypotenuse = 7/25 tanθ = sinθ/cosθ = Opposite side/Adjacent side = 7/24 sin2θ = (7/25)2 = 49/625 cos2θ = (24/25)2 = 576/625 Question 8. Fill in the blanks. i. ii. iii. Answer: i. We know the following identity, sinθ = cos (90 -θ) So sin 20° = cos (90 – 20) ∴ sin 20° = cos 70° ii. We know that, Let the unknown angle be θ tan 30° = tanθ = = tanθ = √3 θ = tan-1(√3) ∴θ = 60° iii. We know that, cosθ = sin (90 -θ ) cos 40° = sin (90 – 40) ∴ cos 40° = sin 50° Page 2
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