In triangle ABC ,if <A and <B are acute angles, and sin A=10/13 what is the value of cos A
To find the value of cos A, we can use the Pythagorean identity: sin^2 A + cos^2 A = 1.
Given that sin A = 10/13, we can substitute this value into the equation:
(10/13)^2 + cos^2 A = 1
Simplifying this equation:
100/169 + cos^2 A = 1
To isolate cos^2 A, we subtract 100/169 from both sides:
cos^2 A = 1 - 100/169
cos^2 A = (169 - 100) / 169
cos^2 A = 69/169
Taking the square root of both sides, we get:
cos A = √(69/169)
Thus, the value of cos A is √(69/169).