In right triangle ABC, with angle a=90, we have AB=6 and BC=10. Find Cos A
B
6 10
A C
Cos A = AC / BC
By the Pythagorean Theorem
AC = sqrt ( BC^2 - AB^2) = sqrt ( 10^2 - 6^2) = sqrt (100 - 36) = sqrt (64) = 8
So
Cos A = 8 / 10 = 4 / 5
+87 
