In triangle ABC if <A and <B are acute angles and sin 10/13 , what is the value of cos A?

To find the value of cos A in triangle ABC, we need to use the given information on sin and the properties of trigonometric ratios.

First, let's draw triangle ABC and label the sides and angles:

A
/ \
c / \ b
/ \
B -----C

Here we have an acute triangle ABC. Let's label the sides as a, b, and c, where side a is opposite angle A, side b is opposite angle B, and side c is opposite angle C.

Given that sin A = 10/13, we can use the Pythagorean identity to find cos A.

The Pythagorean identity in a right triangle is:
sin^2(theta) + cos^2(theta) = 1

Since angle A is an acute angle, we can use the Pythagorean identity in the triangle ABC to find cos A:

sin^2(A) + cos^2(A) = 1

(10/13)^2 + cos^2(A) = 1

Simplifying, we get:
100/169 + cos^2(A) = 1

To solve for cos^2(A), rearrange the equation:
cos^2(A) = 1 - 100/169

cos^2(A) = (169 - 100)/169

cos^2(A) = 69/169

Now, take the square root of both sides to find cos A:
cos(A) = sqrt(69/169)

Therefore, the value of cos A in triangle ABC is sqrt(69/169).