Question

in a right angle triangle ABC right angled at b if 15 cot a = b then find sin a and sec b​

Answer 1

Answer:

15cot a=b

cot a=b/15

1 + cot^2a = 1 +(b/15)^2

cosec^2a = (225+b^2)/225

cosec a = √(225+b^2)/15

sin a = 1/cosec a = 15/√(b^2+225)Ans

sec b = sec90 = infinity Ans