If cos =40/9 find the values of cosec &
sin
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0SaveLet us draw a ∆ ABC in which ∠B = 90°.
0SaveLet us draw a ∆ ABC in which ∠B = 90°.Then cos θ = AB/AC = 9/41.
0SaveLet us draw a ∆ ABC in which ∠B = 90°.Then cos θ = AB/AC = 9/41.Let AB = 9k and AC = 41k, where k is positive.
0SaveLet us draw a ∆ ABC in which ∠B = 90°.Then cos θ = AB/AC = 9/41.Let AB = 9k and AC = 41k, where k is positive.By Pythagoras’ theorem, we get
0SaveLet us draw a ∆ ABC in which ∠B = 90°.Then cos θ = AB/AC = 9/41.Let AB = 9k and AC = 41k, where k is positive.By Pythagoras’ theorem, we getAC2 = AB2 + BC2
0SaveLet us draw a ∆ ABC in which ∠B = 90°.Then cos θ = AB/AC = 9/41.Let AB = 9k and AC = 41k, where k is positive.By Pythagoras’ theorem, we getAC2 = AB2 + BC2⇒ BC2 = AC2 – AB2
0SaveLet us draw a ∆ ABC in which ∠B = 90°.Then cos θ = AB/AC = 9/41.Let AB = 9k and AC = 41k, where k is positive.By Pythagoras’ theorem, we getAC2 = AB2 + BC2⇒ BC2 = AC2 – AB2⇒ BC2 = [(41k)2 – (9k)2]
0SaveLet us draw a ∆ ABC in which ∠B = 90°.Then cos θ = AB/AC = 9/41.Let AB = 9k and AC = 41k, where k is positive.By Pythagoras’ theorem, we getAC2 = AB2 + BC2⇒ BC2 = AC2 – AB2⇒ BC2 = [(41k)2 – (9k)2]⇒ BC2 = [1681k2 – 81k2]
0SaveLet us draw a ∆ ABC in which ∠B = 90°.Then cos θ = AB/AC = 9/41.Let AB = 9k and AC = 41k, where k is positive.By Pythagoras’ theorem, we getAC2 = AB2 + BC2⇒ BC2 = AC2 – AB2⇒ BC2 = [(41k)2 – (9k)2]⇒ BC2 = [1681k2 – 81k2]⇒ BC2 = 1600k2
0SaveLet us draw a ∆ ABC in which ∠B = 90°.Then cos θ = AB/AC = 9/41.Let AB = 9k and AC = 41k, where k is positive.By Pythagoras’ theorem, we getAC2 = AB2 + BC2⇒ BC2 = AC2 – AB2⇒ BC2 = [(41k)2 – (9k)2]⇒ BC2 = [1681k2 – 81k2]⇒ BC2 = 1600k2⇒ BC = √(1600k2)
0SaveLet us draw a ∆ ABC in which ∠B = 90°.Then cos θ = AB/AC = 9/41.Let AB = 9k and AC = 41k, where k is positive.By Pythagoras’ theorem, we getAC2 = AB2 + BC2⇒ BC2 = AC2 – AB2⇒ BC2 = [(41k)2 – (9k)2]⇒ BC2 = [1681k2 – 81k2]⇒ BC2 = 1600k2⇒ BC = √(1600k2)⇒ BC = 40k
0SaveLet us draw a ∆ ABC in which ∠B = 90°.Then cos θ = AB/AC = 9/41.Let AB = 9k and AC = 41k, where k is positive.By Pythagoras’ theorem, we getAC2 = AB2 + BC2⇒ BC2 = AC2 – AB2⇒ BC2 = [(41k)2 – (9k)2]⇒ BC2 = [1681k2 – 81k2]⇒ BC2 = 1600k2⇒ BC = √(1600k2)⇒ BC = 40kTherefore, sin A = BC/AC = 40k/41k = 40/41
0SaveLet us draw a ∆ ABC in which ∠B = 90°.Then cos θ = AB/AC = 9/41.Let AB = 9k and AC = 41k, where k is positive.By Pythagoras’ theorem, we getAC2 = AB2 + BC2⇒ BC2 = AC2 – AB2⇒ BC2 = [(41k)2 – (9k)2]⇒ BC2 = [1681k2 – 81k2]⇒ BC2 = 1600k2⇒ BC = √(1600k2)⇒ BC = 40kTherefore, sin A = BC/AC = 40k/41k = 40/41cos A = AB/AC = = 9k/41k = 9/41
0SaveLet us draw a ∆ ABC in which ∠B = 90°.Then cos θ = AB/AC = 9/41.Let AB = 9k and AC = 41k, where k is positive.By Pythagoras’ theorem, we getAC2 = AB2 + BC2⇒ BC2 = AC2 – AB2⇒ BC2 = [(41k)2 – (9k)2]⇒ BC2 = [1681k2 – 81k2]⇒ BC2 = 1600k2⇒ BC = √(1600k2)⇒ BC = 40kTherefore, sin A = BC/AC = 40k/41k = 40/41cos A = AB/AC = = 9k/41k = 9/41tan A = Sin A/Cos A = (40/41 × 41/9) = 40/9
0SaveLet us draw a ∆ ABC in which ∠B = 90°.Then cos θ = AB/AC = 9/41.Let AB = 9k and AC = 41k, where k is positive.By Pythagoras’ theorem, we getAC2 = AB2 + BC2⇒ BC2 = AC2 – AB2⇒ BC2 = [(41k)2 – (9k)2]⇒ BC2 = [1681k2 – 81k2]⇒ BC2 = 1600k2⇒ BC = √(1600k2)⇒ BC = 40kTherefore, sin A = BC/AC = 40k/41k = 40/41cos A = AB/AC = = 9k/41k = 9/41tan A = Sin A/Cos A = (40/41 × 41/9) = 40/9csc A = 1/sin A = 41/40
0SaveLet us draw a ∆ ABC in which ∠B = 90°.Then cos θ = AB/AC = 9/41.Let AB = 9k and AC = 41k, where k is positive.By Pythagoras’ theorem, we getAC2 = AB2 + BC2⇒ BC2 = AC2 – AB2⇒ BC2 = [(41k)2 – (9k)2]⇒ BC2 = [1681k2 – 81k2]⇒ BC2 = 1600k2⇒ BC = √(1600k2)⇒ BC = 40kTherefore, sin A = BC/AC = 40k/41k = 40/41cos A = AB/AC = = 9k/41k = 9/41tan A = Sin A/Cos A = (40/41 × 41/9) = 40/9csc A = 1/sin A = 41/40sec A = 1/cos A = 41/9 and
0SaveLet us draw a ∆ ABC in which ∠B = 90°.Then cos θ = AB/AC = 9/41.Let AB = 9k and AC = 41k, where k is positive.By Pythagoras’ theorem, we getAC2 = AB2 + BC2⇒ BC2 = AC2 – AB2⇒ BC2 = [(41k)2 – (9k)2]⇒ BC2 = [1681k2 – 81k2]⇒ BC2 = 1600k2⇒ BC = √(1600k2)⇒ BC = 40kTherefore, sin A = BC/AC = 40k/41k = 40/41cos A = AB/AC = = 9k/41k = 9/41tan A = Sin A/Cos A = (40/41 × 41/9) = 40/9csc A = 1/sin A = 41/40sec A = 1/cos A = 41/9 andcot A = 1/tan A = 9/40.
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