Consider the right triangle ABC with legs AB=4, AC=3 and hypotenuse BC=5. Angle B has
[tex]\sin B=\frac{opposite}{hypotenuse} = \frac{3}{5} [/tex] and
[tex]\cos B = \frac{adjacent}{hypotenuse} = \frac{4}{5} [/tex].
Since O lies in second quadrant [tex]\cos O\ \textless \ 0 [/tex] and
[tex]\cos O= -\cos B =- \frac{4}{5} [/tex].
Answer: [tex]\cos O=- \frac{4}{5} [/tex].
