Question

Q. If cot A = k, then sin A is equal to
(presume that A is an acute angle)​

Answer 1

Answer:

SOLUTION

SOLUTIONGiven:

SOLUTIONGiven:cot A = k

SOLUTIONGiven:cot A = kFormula:

SOLUTIONGiven:cot A = kFormula:cot A = Base/Perpendicular

SOLUTIONGiven:cot A = kFormula:cot A = Base/Perpendicularsin A = Perpendicular/Hypotenuse

SOLUTIONGiven:cot A = kFormula:cot A = Base/Perpendicularsin A = Perpendicular/Hypotenuse (Hypotenuse)2 = (Base)2 + (Perpendicular)2

SOLUTIONGiven:cot A = kFormula:cot A = Base/Perpendicularsin A = Perpendicular/Hypotenuse (Hypotenuse)2 = (Base)2 + (Perpendicular)2Calculation:

SOLUTIONGiven:cot A = kFormula:cot A = Base/Perpendicularsin A = Perpendicular/Hypotenuse (Hypotenuse)2 = (Base)2 + (Perpendicular)2Calculation:cot A = k/1

SOLUTIONGiven:cot A = kFormula:cot A = Base/Perpendicularsin A = Perpendicular/Hypotenuse (Hypotenuse)2 = (Base)2 + (Perpendicular)2Calculation:cot A = k/1⇒ Base/Perpendicular = k/1

SOLUTIONGiven:cot A = kFormula:cot A = Base/Perpendicularsin A = Perpendicular/Hypotenuse (Hypotenuse)2 = (Base)2 + (Perpendicular)2Calculation:cot A = k/1⇒ Base/Perpendicular = k/1⇒ Base = k and Perpendicular = 1

SOLUTIONGiven:cot A = kFormula:cot A = Base/Perpendicularsin A = Perpendicular/Hypotenuse (Hypotenuse)2 = (Base)2 + (Perpendicular)2Calculation:cot A = k/1⇒ Base/Perpendicular = k/1⇒ Base = k and Perpendicular = 1(Hypotenuse)2 = (Base)2 + (Perpendicular)2

SOLUTIONGiven:cot A = kFormula:cot A = Base/Perpendicularsin A = Perpendicular/Hypotenuse (Hypotenuse)2 = (Base)2 + (Perpendicular)2Calculation:cot A = k/1⇒ Base/Perpendicular = k/1⇒ Base = k and Perpendicular = 1(Hypotenuse)2 = (Base)2 + (Perpendicular)2⇒ (Hypotenuse)2 = k2 + 12

SOLUTIONGiven:cot A = kFormula:cot A = Base/Perpendicularsin A = Perpendicular/Hypotenuse (Hypotenuse)2 = (Base)2 + (Perpendicular)2Calculation:cot A = k/1⇒ Base/Perpendicular = k/1⇒ Base = k and Perpendicular = 1(Hypotenuse)2 = (Base)2 + (Perpendicular)2⇒ (Hypotenuse)2 = k2 + 12⇒ Hypotenuse = √(1 + K2)

SOLUTIONGiven:cot A = kFormula:cot A = Base/Perpendicularsin A = Perpendicular/Hypotenuse (Hypotenuse)2 = (Base)2 + (Perpendicular)2Calculation:cot A = k/1⇒ Base/Perpendicular = k/1⇒ Base = k and Perpendicular = 1(Hypotenuse)2 = (Base)2 + (Perpendicular)2⇒ (Hypotenuse)2 = k2 + 12⇒ Hypotenuse = √(1 + K2)Now, sin A = 1/√(1 + k2)