1. Check the picture below.
2. Tan∅=8/3 so let's construct the right triangle ABC with BC=8, BA=3, m(B)=90° and m(A) =∅.
3. By Pythagorean theorem, AC=[tex] \sqrt{ 8^{2} + 3^{2} } = \sqrt{64+9} = \sqrt{73} [/tex]
4. cos∅=adjecent side/hypothenuse=[tex] \frac{ 3}{ \sqrt{73}}= \frac{ 3}{ \sqrt{73}} \frac{\sqrt{73}}{\sqrt{73}}= \frac{3\sqrt{73}}{73} [/tex]
5. Cos∅ is negative so ∅ is in an angle in the second or third quadrant. The assumption that ∅ is and acute angle of a right triangle only makes the solution more practical. The only thing that changes, is sign, which depends on the quadrant.
6. As cos ∅ is negative, cos∅=-[tex]\frac{3\sqrt{73}}{73} [/tex] .
