Question 1193960

For positive acute angles A and B,, it is known that sin A=9/41 and cos B=3/5 

​
 Find the value of cos (A-B) in simplest form. I know that sin A =9/41, 40/41 by drawing a triangle for help me to understand how to set up.
 Robbin
<pre>cos (A - B) = cos A cos B + sin A sin B

Based on the above, you'll need cos A and sin B, since cos B and sin A were given.

          As {{{matrix(1,7, sin (A), "=", 9/41, "=", O/H, "=", y/r)}}} we can see that x, or the adjacent (A) side MUST be 40 since this is a 9-40-41 Pythagorean Triple. 
We then get: {{{matrix(1,7, cos (A), "=", A/H, "=", x/r, "=", 40/41)}}}

          As {{{matrix(1,7, cos (B), "=", 3/5, "=", A/H, "=", x/r)}}} we can see that y, or the opposite (O) side MUST be 4 since this is a 3-4-5 Pythagorean Triple. 
We then get: {{{matrix(1,7, sin (B), "=", O/H, "=", y/r, "=", 4/5)}}}

          {{{matrix(1,3, cos (A - B), "=", cos (A) cos (B) + sin (A) sin (B))}}}
                   {{{highlight_green(matrix(3,2, "=", (40/41)(3/5) + (9/41)(4/5),
"=", 120/205 + 36/205, "=", highlight(156/205)))}}}</pre>