Science:Math Exam Resources/Courses/MATH102/December 2013/Question A 02/Solution 1

Let $\theta =\arcsin \left({\frac {\sqrt {9-x^{2}}}{3}}\right)$. Then $\sin(\theta )={\frac {\sqrt {9-x^{2}}}{3}}$. This gives the following triangle:

To solve for the last side, we simply use the Pythagorean theorem: $a^{2}=3^{2}-\left({\sqrt {9-x^{2}}}\right)^{2}=9-9+x^{2}=x^{2}$ and so $a=x$ (we only need to consider the positive root since $a$ has to be positive).

Hence, taking the cosine of theta in our triangle above, we see that $\cos(\theta )=\cos \left(\arcsin \left({\frac {\sqrt {9-x^{2}}}{3}}\right)\right)={\frac {a}{3}}={\frac {x}{3}}$, which is answer ${\color {blue}{\text{(B)}}}$.