If sin A =3/4, calculate cos A and tan A.
Let ΔABC be a right-angled triangle, right-angled at B.
We know that sin A = BC/AC = 3/4
Let BC be 3k and AC will be 4k where k is a positive real number.
By Pythagoras theorem we get,
AC2 = AB2 + BC2
(4k)2 = AB2 + (3k)2
16k2 - 9k2 = AB2
AB2 = 7k2
AB = √7 k
cos A = AB/AC = √7 k/4k = √7/4
tan A = BC/AB = 3k/√7 k = 3/√7
