<PRE><B>Question 30116</B>
In triangle AB, A and B are acute angles so. AC and BC are right angles.
Sin A = 10/13
13 is the hypotenuse side and 10 is the opposite side.
To find cos A.
From pythagorean thereom:

*[tex a^2+b^2=c^2]

*[tex 10^2+b^2=13^2]

*[tex b=\sqrt{69}]

b=8.3


SO the value for cos A is 8.3/13
cos A=0.63 take the inverse sine of cosine:
cos A=51 degrees.
PAul.</PRE>