Given that for positive acute angles A and B:
![\begin{gathered} \cos A=(60)/(61) \\ \\ \tan B=(12)/(35) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/41ukt85ubgv95gxi63u32poslqe2j0t91b.png)
You need to remember that, by definition:
![\begin{gathered} \cos \alpha=(adjacent)/(hypotenuse) \\ \\ \tan \alpha=(opposite)/(adjacent) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/u786l8bndig9uhfettjo0qjlxdor0rh5ga.png)
Look at the Right Triangle shown below:
It is also important to remember the following Trigonometric Identity:
![\cos \mleft(A-B\mright)=cosA\cdot cosB+sinA\cdot sinB](https://img.qammunity.org/2023/formulas/mathematics/college/cuerg1duizf1b9yhzdnrgzgstl0q0x64su.png)
Then, in order to solve this exercise, you need to find:
![\begin{gathered} \cos B \\ \sin A \\ \sin B \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/sg2312vikcno4uycbvmu9gcin7txhkk0vq.png)
Using the data given in the exercise, you can draw these two Right Triangles (they are not drawn to scale):
Since by definition:
![\sin \alpha=(opposite)/(hypotenuse)](https://img.qammunity.org/2023/formulas/mathematics/college/np687vjr2n6ghtkhhi6tl9n2e8aaavlj97.png)
You need to find the opposite side of the first triangle and the hypotenuse of the second triangle. You can do this by applying the Pythagorean Theorem, which states that:
![c^2=a^2+b^2](https://img.qammunity.org/2023/formulas/mathematics/high-school/udh1dsx7kwgfauditnn86pp2qhoycm1tvv.png)
Where "c" is the hypotenuse, and "a" and "b" are the legs of the Right Triangle.
Then, the opposite side of the first triangle is:
![\begin{gathered} 61^2=60^2+b^2 \\ \\ 61^2-60^2=b^2 \\ \\ \sqrt[]{61^2-60^2}=b \\ \\ b=11 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/4c2m8xf45vf5n3a4yocre3wocemgz5jt4s.png)
And the hypotenuse of the second triangle is:
![\begin{gathered} c^2=12^2+35^2 \\ \\ c=\sqrt[]{12^2+35^2} \\ \\ c=37 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/2a289pwp6099ivwaea5xx5enco8ouamrwj.png)
Now you can determine that:
![\begin{gathered} \cos B=(35)/(37) \\ \\ \sin A=(11)/(61) \\ \\ \sin B=(12)/(37) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/zx3rojk55pxy0lvpgnsgtroii8fzfltfj5.png)
Substitute values into the Trigonometric Identity:
![\begin{gathered} \cos (A-B)=cosA\cdot cosB+sinA\cdot sinB \\ \\ \cos (A-B)=((60)/(61))((35)/(37))+((11)/(61))((12)/(37)) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ma2hge6fmqhqxpihzsqd362ky79revupm3.png)
Simplify:
- Multiply the fractions:
![\begin{gathered} \cos (A-B)=(60\cdot35)/(61\cdot37)+(11\cdot12)/(61\cdot37) \\ \\ \cos (A-B)=(2100)/(2257)+(132)/(2257) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/78h7dndv2y4z4e31tmw7bzscqsotwd2jtv.png)
- Add the fractions:
![\begin{gathered} \cos (A-B)=(2100+132)/(2257) \\ \\ \cos (A-B)=(2232)/(2257) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/6561k6gkc8tcto0cdo2xu6p035fouv2cz0.png)
Hence, the answer is:
![\cos (A-B)=(2232)/(2257)](https://img.qammunity.org/2023/formulas/mathematics/college/3f7m8z2y2z32rum88arhwvamwybxm19347.png)