In right triangle $ABC$ above, $AB=10$. If the cosine of $x \degree$ is $\dfrac{\sqrt{3}}{2}$, what is the length of $\overline{AC}$?

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Since cosine refers to the ratio of the adjacent side to the hypotenuse,

$$\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{\sqrt{3}}{2}$$

These are the ratios for a a 30-60-90 triangle.

These are the ratios for a a 30-60-90 triangle. The hypotenuse for our triangle has a hypotenuse of 10, so $\overline{AC}$ should be half the length, or $\boxed{5}$.

Similar to the previous approach, we can use inverse trigonometric functions to find the measure of $x$.

$$\cos{x}=\frac{\sqrt{3}}{2}$$ $$x=\cos^{-1}{\left(\frac{3}{2}\right)}$$ $$x= 30\degree$$

We can then follow the strategy for a 30-60-90 triangle as shown in approach 1.

Since cosine refers to the ratio of the adjacent side to the hypotenuse,

$$\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{\sqrt{3}}{2}$$

We can use the pythagorean theorem to obtain the opposite side.

$$a^2+b^2=c^2$$ $$(\sqrt{3})^2+b^2=2^2$$ $$3+b^2=4$$ $$b^2=1$$ $$b=1$$

Therefore, the ratio of the opposite side to the hypotenuse is 1:2. Since our given triangle has a hypotenuse of 10, the corresponding side opposite of angle $x$ will be $\boxed{5}$.