397 (a) Here, cosec2 A – cot2 A = 1 LHS = cosec2 A – cot2 A = h p 2 – b p 2 = 5 4 2 – 3 4 2 = 25 16 – 9 16 = 25 – 9 16 = 16 16 = 1 = RHS, Proved. (b) Here, 1 – cos2 A = sin2 A LSH = 1 – cos2 A = 1 – b h 2 = 1 – 3 5 2 = 1 – 9 25 = 25 – 9 25 = 16 25 RHS = sin2 A = p h 2 = 4 5 2 = 16 25 LHS = RHS, Proved. (c) Here, 1 + tan2 A = sec2 A LHS = 1 + tan2 A = 1 + p b 2 = 1 + 4 3 2 = 1 + 16 9 = 9 + 16 9 = 25 16 RHS = sec2 A = h p 2 = 5 4 2 = 25 16 LHS = RHS, Proved. 8. If tan A = 5 13 , prove that: (a) sec A. cos A = 1 (b) cosec2 A – cot2 A = 1 (c) sin2 A + cos2 A = 1 Solution: Here, tan A = 5 13 i.e. p b = 5 13 p = 5, b = 13 by Pythagoras theorem, h2 = p2 + b2 = 52 + 132 = 25 + 169 = 194 or, h = 194 (a) Here, sec A . cos A = 1 LHS = sec A . cos A = h b × b h = 1 = RHS, Proved. (b) Here, cosec2 A – cot2 A = 1 LHS = h p 2 – b p 2 = 194 5 2 – 13 5 2 = 194 25 – 169 25 = 194 – 169 25 = 25 25 = 1 = RHS, Proved.
398 (c) Here, sin2 A + cos2 A = 1 LHS = p h 2 + b h 2 = 5 194 2 + 13 194 2 = 25 194 + 169 194 = 25 + 169 194 = 194 194 = 1 = RHS, Proved. 9. From the following figures, find the values of sin θ, cos θ and tan θ: (a) (b) (c) (d) Solution: (a) Here, in right angled triangle BCD BC2 = CD2 + BD2 [ by Pythagoras theorem] or, BC2 = (4 cm)2 + (3 cm)2 = 16 cm + 9 cm = 25 cm or, BC = 5 cm Now, in right angled triangle ABC. AB2 = BC2 + CA2 or, AB2 = (5 cm)2 + (12 cm)2 = 25 cm2 + 144 cm2 = 169 cm2 or, AB = 13 cm Hence, for reference angle , p = BC = 5 cm b = CA = 12 cm h = AB = 13 cm Finally, sin = p h = 5 13 cos = b h = 12 13 tan = p b = 5 12 (b) In right angled triangle PQR, by Pythagoras theorem, PR2 = PQ2 + QR2 or, PR2 = (8 cm)2 + (6 cm)2 = 64 cm2 + 36 cm2 = 100 cm2 PR = 10 cm Again, in right angled triangle PRS by Pythagoras theorem, PS2 = PR2 + RS2 or, (26 cm)2 = (10 cm)2 + RS2
399 or, 676 cm2 = 100 cm2 + RS2 or, 576 cm2 = RS2 RS = 24 cm Now, for reference angle , p = PR = 10 cm b = RS = 24 cm h = PS = 26 cm Hence, sin = p h = 5 13 cos = b h = 12 13 tan = p b = 5 12 (c) In right angled triangle ACD, by Pythagoras theorem, AC2 = AD2 + DC2 or, AC2 = (7 cm)2 + (24 cm)2 = 49 cm2 + 576 cm2 = 625 cm2 AC = 25 cm Again, in right angled triangle ABC, AC2 = AB2 + BC2 or, (25 cm)2 = (20 cm)2 + BC2 or, 625 cm2 = 400 cm2 + BC2 or, 225 cm2 = BC2 BC = 15 cm Now, for reference angle , p = BC = 15 cm b = AB = 20 cm h = AC = 25 cm Hence, sin = p h = 15 25 = 3 5 cos = b h = 20 25 = 4 5 tan = p b = 15 20 = 3 4 (d) In right angled triangle LMK, by Pythagoras theorem, KM2 = LM2 + KL2 or, (29 cm)2 = LM2 + (21 cm)2 or, 841 cm2 = LM2 + 441 cm2 or, LM2 = 400 cm2 or, LM = 20 cm Again, in right angled triangle LMN, LN2 = LM2 + MN2 or, LN2 = (20 cm)2 + (15 cm)2 = 400 cm2 + 225 cm2 = 625 cm2 LN = 25 cm
400 Now, for reference angle , p = MN = 15 cm b = LM = 20 cm h = LN = 25 cm Hence, sin = p h = 15 25 = 3 5 cos = b h = 20 25 = 4 5 tan = p b = 15 20 = 3 4 10. From the following figures, find the values of sin θ, cos θ and tan θ: (a) (b) Solution: (a) In right angled triangle ABD, by Pythagoras theorem, AB2 = AD2 + DB2 or, (4 cm)2 = (3 cm)2 + DB2 or, 16 cm2 = 9 cm2 + DB2 or, DB2 = 7 cm2 or, DB = 7 cm Again, in right angled triangle BCD. by Pythagoras theorem, BC2 = BD2 + DC2 or, (3 cm)2 = ( 7 cm)2 = DC2 or, 9 cm2 = 7 cm2 + DC2 or, DC2 = 2 cm2 or, DC = 2 cm Now, for reference angle , p = DC = 2 cm b = BD = 7 cm h = BC = 3 cm Hence, sin = p h = 2 3 cos = b h = 7 3 tan = p b = 2 7 = 2 7
401 (b) In right angled triangle PQR, PR = PT + TQ = 12 cm + 3 cm = 15 cm by Pythagoras theorem, PR2 = PQ2 + QR2 or, (15 cm)2 = (12 cm)2 + QR2 or, 225 cm2 = 144 cm2 + QR2 or, QR2 = 81 cm2 or, QR = 9 cm In right angled triangle TSR SR = TQ – QS = 9 cm – 4 cm = 5 cm by Pythagoras theorem, SR2 = ST2 + TR2 or, (5 cm)2 = ST2 + (3 cm)2 or, 25 cm2 = ST2 + 9 cm2 or, ST2 = 16 cm2 or, ST = 4 cm Now, for reference angle , p = TR = 3 cm b = ST = 4 cm h = SR = 5 cm Hence, sin = p h = 3 5 cos = b h = 4 5 tan = p b = 3 4 11. (a) Express sin θ into cos θ. (b) Express sin A into tan A. Solution: (a) Here, sin = p h = h2 – b2 h [ h2 = p2 + b2 ] = h2 – b2 h2 = h2 h2 – b2 h2 = 1 – b h 2 = 1 – cos2 (b) Here, sin A = p h = p b h b = p b p2 + b2 b [ h2 = p2 + b2 ] = tan A p2 + b2 b2 = tan A p2 b2 + b2 b2 = tan A p b 2 + 1 = tan A tan2 A + 1
402 12. (a) A vertical pole of 16 ft is supported by a wire of length 20 ft. What is the distance between the bottom of the pole and base of wire? If the wire makes an angle θ with the ground, find the values of sin θ, cos θ and tan θ. (b) A 25 m tall tower castes 60 m long shadow. How long wire will need to support the tower? If the wire makes an angle with the ground, find the values of sin θ, sec θ and cot θ. Solution: (a) Here, let AB = 10 ft be the height of pole and BC = 20 ft the length of wire To find: distance between the bottom of pole and base of wire (CA) = ? In right angled triangle ABC, by Pythagoras theorem, BC2 = AB2 + CA2 or, (20 ft)2 = (16 ft)2 + CA2 or, 400 ft2 = 256 ft2 = CA2 or, 144 ft2 = CA2 or, CA = 12 ft Now, for reference angle , p = AB = 16 ft b = CA = 12 ft h = BC = 20 ft Hence, sin = p h = 16 20 = 4 5 cos = b h = 12 20 = 3 4 tan = p b = 16 12 = 4 3 (b) Let, AB = 25 m be the height of tower and AC = 60 m be the shadow casted by tower. To find: length of wire need to support tower (BC) = ? In right angled triangle ABC, by Pythagoras theorem, BC2 = AB2 + AC2 or, BC2 = (25 m)2 + (60 m)2 = 625 m2 + 3600 m2 = 4225 m2 or, BC = 65 m Now, for the reference angle , p = AB = 25 m b = AC = 60 cm h = BC = 65 cm Hence, sin = p h = 25 65 = 5 13 cos = b h = 60 65 = 12 13 tan = p b = 25 60 = 5 12
403 20.2 Trigonometric Ratios of Standard Angles PRACTICE 20.2 Read / Understand / Think / Do Keeping Skill Sharp 1. Without using tables or calculator, evaluate: (a) sin 30° + cos 30° + tan 30° (b) sin 0° + cos 0° + tan 0° (c) 2sin 30°. cos 30° (d) 2 tan 30° 1 – tan2 30° (e) sin 30°.cos 45° + cos 30°.sin 45° (f) cos 30°.cos 45° – sin 30°.sin 45° Solution: (a) sin 30° + cos 30° + tan 30° = 1 2 + 3 2 + 1 3 = 3 + 3 + 2 2 3 = 5 + 3 2 3 = 5 + 3 2 3 × 3 3 = 5 3 + 3 2 × 3 = 3 + 5 3 6 (b) sin 0° + cos 0° + tan 0° = 0 + 1 + 0 = 1 (c) 2sin 30°. cos 30° = 2 × 1 2 × 3 2 = 3 2 (d) 2 tan 30° 1 – tan2 30° = 2 × 1 3 1 – 1 3 2 = 2 3 1 – 1 3 = 2 3 3 – 1 3 = 2 3 × 3 2 = 3 3 × 3 3 = 3 3 3 = 3 (e) sin 30°.cos 45° + cos 30°.sin 45° = 1 2 × 1 2 + 3 2 × 1 2 = 1 2 2 + 3 2 2 = 1 + 3 2 2
404 = 1 + 3 2 2 × 2 2 = 2 + 6 2 × 2 = 2 + 6 4 (f) cos 30°.cos 45° – sin 30°.sin 45° = 3 2 × 1 2 – 1 2 × 1 2 = 3 2 2 – 1 2 2 = 3 – 1 2 2 = 3 – 1 2 2 × 2 2 = 6 – 2 2 × 2 = 6 – 2 4 2. Prove the following relations: (a) sin 45°.cos 30° – cos 45°. sin 45° = 6 – 2 4 (b) sin 60°.cos 30° – cos 60°. sin 30° = 1 2 (c) 2 tan 30° 1 – tan2 30° = 3 (d) cos 60° = 2cos2 30° – 1 (e) tan 60° – tan 30° 1 + tan 60° . tan 30° = 1 3 (f) sin 30°.cos 60° + cos 30°. sin 60° = sin 90° Solution: (a) Here, sin 45°.cos 30° – cos 45°. sin 45° = 6 – 2 4 LHS = sin 45°.cos 30° – cos 45°. sin 45° = 1 2 × 3 2 – 1 2 × 1 2 = 3 2 2 – 1 2 2 = 3 – 1 2 2 × 2 2 = 6 – 2 2 × 2 = 6 – 2 4 = RHS, Proved.
405 (b) Here, sin 60°.cos 30° – cos 60°. sin 30° = 1 2 LHS = sin 60°.cos 30° – cos 60°. sin 30° = 3 2 × 3 2 – 1 2 × 1 2 = 3 4 – 1 4 = 3 – 1 4 = 2 4 = 1 2 = RHS, Proved. (c) Here, 2 tan 30° 1 – tan2 30° = 3 LHS = 2 tan 30° 1 – tan2 30° = 2 × 1 3 1 – 1 3 2 = 2 3 1 – 1 3 = 2 3 3 – 1 3 = 2 3 2 3 = 2 3 × 3 2 = 3 3 × 3 3 = 3 = RHS, Proved. (d) Here, cos 60° = 2cos2 30° – 1 LHS = cos 60° = 1 2 RHS = 2cos2 30° – 1 = 2 × 3 2 2 – 1= 2 × 3 4 – 1 = 3 2 – 1 = 3 – 2 2 = 1 2 LHS = RHS, Proved. (e) Here, tan 60° – tan 30° 1 + tan 60° . tan 30° = 1 3 LHS = tan 60° – tan 30° 1 + tan 60° . tan 30° = 3 – 1 3 1 + 3 × 1 3 = 3 – 1 3 1 + 1 = 2 3 2 = 2 3 × 1 2 = 1 3 = RHS, Proved.
406 (f) Here, sin 30°.cos 60° + cos 30°. sin 60° = sin 90° LHS = sin 30°.cos 60° + cos 30°. sin 60° = 1 2 × 1 2 + 3 2 × 3 2 = 1 4 + 3 4 = 4 4 = 1 RHS = sin 90° = 1 LHS = RHS, Proved. 3. (a) If sin θ = 1 2 , find cos θ and tan θ. (b) If 2sin θ = 3 , find cos θ and tan θ. (c) If cos θ = 1, find sin θ and tan θ. (d) If sin θ = cos θ, find sin θ and cos θ. Solution: (a) Here, sin = 1 2 (b) Here, 2 sin = 3 i.e. sin = sin 30° i.e. sin = 3 2 = 30° or, sin = sin 60° So, cos = cos 30° = 3 2 = 60° tan = tan 30° = 1 3 So, cos = cos 60° = 1 2 tan = tan 60° = 3 (c) Here, cos = 1 (d) Here, sin = cos i.e. cos = cos 0° or, sin cos = 1 = 0° or, tan = 1 So, sin = sin 0° = 0 or, tan = tan 45° tan = tan 0° = 0 = 45° So, sin = sin 45° = 1 2 cos = cos 45°= 1 2 4. Verify that: (a) tan 60° = 2 tan 30° 1 – tan2 30° (b) cos 60° = 2 cos2 30° – 1 (c) cos 90° = cos 30°.cos 60° – sin 30°.sin 60° (d) 1 – sin 60° cos 60° = – 1 – tan 30° 1 + tan 30° Solution: (a) Here, tan 60° = 2 tan 30° 1 – tan2 30° LHS = tan 60° = 3
407 RHS = 2 tan 30° 1 – tan2 30° = 2 × 1 3 1 – 1 3 2 = 2 3 1 – 1 3 = 2 3 3 – 1 3 = 2 3 2 3 = 2 3 × 3 2 = 3 3 × 3 3 = 3 3 3 = 3 = RHS, Proved. (b) Here, cos 60° = 2 cos2 30° – 1 LHS = cos 60° = 1 2 RHS = 2 cos2 30° – 1 = 2 × 3 2 2 – 1 = 2 × 3 4 – 1 = 3 2 – 1 = 3 – 2 2 = 1 2 LHS = RHS, Proved. (c) Here, cos 90° = cos 30°.cos 60° – sin 30°.sin 60° LHS = cos 90° = 0 RHS = cos 30°.cos 60° – sin 30°.sin 60° = 3 2 × 1 2 – 1 2 × 3 2 = 3 4 – 3 4 = 0 LHS = RHS, Proved. (d) Here, 1 – sin 60° cos 60° = – 1 – tan 30° 1 + tan 30° LHS = 1 – sin 60° cos 60° = 1 – 3 2 1 2 = 2 – 3 2 × 2 1 = 2 – 3 RHS = – 1 – tan 30° 1 + tan 30° = – 1 – 1 3 1 + 1 3 = – 3 – 1 3 3 + 1 3 = – 3 – 1 3 × 3 3 + 1 = – 3 – 1 3 + 1 × 3 – 1 3 – 1 = – ( 3 – 1)2 ( 3)2 – 1 = – 3 – 2 3 + 1 3 – 1 = – 4 – 2 3 2 = – 2 3 – 4 2 = – 2( 3 – 2) 2 = 2 – 3 LHS = RHS, Proved. 5. (a) The hypotenuse of a right-angled triangle is 8 cm and one of its side angles is 60°. Find the remaining angle and sides. (b) The hypotenuse of a right-angled triangle is 8 2 cm and one of its side angles is 45°. Find the remaining angle and sides. Solution: (a) Let ABC be the right angled triangle, BC = 8 cm be hypotenuse and ABC = 60°. Hence for reference angle ABC = 60° p = AC b = AB h = BC = 8 cm Hence, sin 60° = p h
408 or, 3 2 = AC 8 or, AC = 3 2 × 8 AC = 4 3 cm Similarly, cos 60°= b h or, 1 2 = AB 8 or, AB = 1 2 × 8 AB = 4 cm Finally, ABC + BCA + CAB = 180° [The sum of angles of triangle] or, 60° + BCA + 90° = 180° or, BCA = 180°– 150° BCA = 30°. (b) Let ABC is be the right angled triangle, BC = 8 2 cm be the hypotenuse and ABC = 45°. Hence, for reference angle ABC = 45° p = AC b = AB h = BC = 8 2 Hence, sin 45° = p h or, 1 2 = AC 8 2 or, AC = 1 2 × 8 2 AC = 8 cm Again, cos 45° = b h or, 1 2 = AB 8 2 or, AB = 1 2 × 8 2 AB = 8 cm Finally, ABC + BCA + CAB = 180° or, 45° + BCA + 90° = 180° or, BCA + 135° = 180° or, BCA = 180° – 135° BCA = 45°.
409 Additional Practice – VII 1. In the given, figure ABC = ACD = 90°, ACB = α and ADC = θ. Also, AD = 13 cm, BC = 4 cm and CD = 12 cm is given. (a) Which trigonometric ratio is equal with AC AB ? (b) Find the length of AC. (c) Find the value of sin q. (d) If ABC is a right-angled isosceles triangle, what would be the value of α in degree? Write with reason. Solution: Here, ABC and ACD are right angled triangles. ABC = ACD = 90° ACB = ADC = AD = 13 cm BC = 4 cm CD = 12 cm (a) AC AB is equal to cosec (b) In right angled triangle ACD by Pythagoras theorem, AD2 = AC2 + DC2 or, (13 cm)2 = AC2 + (12 cm)2 or, 169 cm2 = AC2 + 144 cm2 or, AC2 = 25 cm2 or, AC = 5 cm (c) For reference angle , p = AC = 5 cm, h = DC = 12 cm, h =AD = 13 cm So, sin = p h = 5 13 (d) If ABC is right angled isosceles triangle, the value of must be 45° as the angles sum upto 180° in triangle and isosceles triangle has two equal base angles with remaining 90°. 2. In the given, figure ACB = BDC = 90° and CAB = θ. Also, BD = 3 cm, CD = 4 cm and AC = 12 cm is given. (a) Which trigonometric ratio is equal with BC AB ? (b) Find the length of BC. (c) Find the value of tan θ. (d) Is the value of sin θ = 5 13 ? Why? Solution: Here, ABC and BCD are right angled triangles ACB = BDC = 90° CAB = BD = 3 cm
410 CD = 4 cm AC = 12 cm (a) BC AB is equal to sin . In right angled triangle BCD, by Pythagoras theorem, BC2 = BD2 + DC2 or, BC2 = (3 cm)2 + (4 cm)2 = 9 cm2 + 16 cm2 = 25 cm2 or, BC = 5 cm (c) For reference angle , p = BC = 5 cm, b = AC = 12 cm, h = AB Now, tan = p b = 5 12 (d) by Pythagoras theorem, AB2 = BC2 + AC2 or, AB2 = (5 cm)2 + (12 cm)2 = 25 cm2 + 144 cm2 = 169 cm2 or, AB = 13 cm Hence, sin = p h = 5 13 Yes, the value of sin = 5 13 because the sin is ratio of p and h where p = 5 cm and h = 13 cm. 3. The hypotenuse of a right-angled triangle is 8 cm and one of its acute angle is 30°. (a) Write the relation of Pythagoras theorem. (b) Find the length of BC. (c) Prove that sin2 30° + cos2 30° = 1 (d) For what angle of ABC, AC is equal to BC? Solution: Here, in right angled triangle ABC, For reference angle 30°, h = AB = 8 cm, p = AC, b = BC (a) The Pythagoras theorem is h2 = p2 + b2 or, AB2 = AC2 + BC2 (b) we know that, cos = b h or, cos 30° = BC AC or, 3 2 = BC 8 or, BC = 3 2 × 8 = 4 3 cm (c) Here, LHS = sin2 30° + cos2 30° = 1 2 2 + 3 2 2
411 = 1 4 + 3 4 = 1 + 3 4 = 4 4 = 1 = RHS, Proved. (d) when AC = BC then, ABC = p h = AC BC = AC AC = 1 tan ABC = tan 45° ABC = 45° Hence, for ABC = 45°, AC is equal to BC. 4. In the right-angled triangle ACB, the value of sin θ is 3 6 . (a) Write the trigonometric ratio of sin θ. (b) Find the value of cos θ. (c) Is the opposite side of reference angle shorter than other? Give reason. (d) Prove that: sin2 θ + cos2 θ = 1. Solution: In right angled triangle ABC. For reference angle , h = AB = 6 cm, p = AC = 3 cm, b = BC (a) sin = p h (b) by Pythagoras theorem, AB2 = AC2 + BC2 or, (6 cm)2 = (3 cm)2 + BC2 or, 36 cm2 = 9 cm2 + BC2 or, BC2 = 27 cm2 or, BC = 3 3 cm Hence, cos = b h = BC AB = 3 3 6 = 3 2 (c) Since, cos = 3 2 = cos 30° = 30° Clearly the another angle of right angled triangle ABC is 60°. As reference angle = 30° is the smallest angle among all three, the side opposite to it is also shortest among all three sides. (d) LHS = sin2 + cos2 = sin2 30° + cos2 30° = 1 2 2 + 3 2 2 = 1 4 + 3 4 = 1 + 3 4 = 4 4 = 1 = RHS, Proved.
412 5. In a right-angled triangle PQR, the PRQ = α is the referenced angle. (a) Which trigonometric ratio is denoted by the ratio of PQ PR ? (b) What is the value of PQ PR 2 + QR PR 2 ? Find it. (c) If α = 45° and PR = 10 unit, find the length of base RQ. (d) If PR = 2PQ, what would be the value of α in degree? Solution: In right angled triangle PQR, For reference angle PRQ = h = PR, p = PQ, b = QR (a) The ratio PQ PR is p h which denotes sin . (b) Here, PQ PR 2 + QR PR 2 = PQ2 PR2 + QR2 PQ2 = PQ2 + QR2 PR2 we know that by Pythagoras theorem PR2 = PQ2 + QR2 Hence, PQ2 + QR2 PR2 = PR2 PR2 = 1. (c) If = 45°, PR = 10, RQ = ? cos = b h or, cos 45° = RQ PR or, 1 2 = RQ 10 or, RQ = 10 2 × 2 2 = 10 2 2 = 5 2 units. (d) If PR = 2PQ, sin = p h = PQ PR = PQ 2PQ = 1 2 = sin 30° = 30° Hence for = 30°, PR = 2PQ.