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

Summary Submit Skip Question

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 \(\boxed{5}\).