Question

If in a AABC, right angled at C, tan B = 12, then find sin B

Answer 1

Answer:

sin B = 12/√145

step by step explanation:

tan B = 12/ 1

tan B = 12/ 1 perpendicular / Base = 12 / 1

Sin B = perpendicular/ hypotenuse

we have to find hypotenuse,

H^2 = P^2 + b ^ 2

H^2 = (12)^ 2 + (1)^2

H^2 = 144 + 1

H^2 = 145

H = √145

SinB = 12 / √145