If `sin A =3/4` , calculate cos A and tan A.
Let ΔABC be a right-angled triangle, right-angled at point B.
sin A = 3/4
`(BC)/(AC) = 3/4`
Let BC be 3k. Therefore, AC will be 4k, where k is a positive integer.
Applying Pythagoras theorem in ΔABC, we obtain
AC2 = AB2 + BC2
(4k)2 = AB2 + (3k)2
16k 2 − 9k 2 = AB2
7k 2 = AB2
AB = sqrt7k
`cos A = ("Side adjacent to"angleA)/"Hypotenuse"`
=`(AB)/(AC) = sqrt(7k)/(4k) = sqrt7/4`
`tan A = ("Side adjacent to"angleA)/("Side adjacent to"angleA)`
`= (BC)/(AB) = (3k)/(sqrt7k) = 3/sqrt7`