Question

Find the longer leg of the triangle.

Answer 1

Answer:

A

Step-by-step explanation:

Since the triangle is right use the cosine ratio

cos30° = $\frac{adjacent }{hypotenuse}$ = $\frac{adj}{2\sqrt{3} }$, so

$\frac{\sqrt{3} }{2}$ = $\frac{adj}{2\sqrt{3} }$

Multiply both sides by 2$\sqrt{3}$

adj = $\frac{\sqrt{3} }{2}$ × 2$\sqrt{3}$ = 3

Answer 2

Answer:

Choice A. 3.

Step-by-step explanation:

The triangle in question is a right triangle.

• The length of the hypotenuse (the side opposite to the right angle) is given.
• The measure of one of the acute angle is also given.

As a result, the length of both legs can be found directly using the sine function and the cosine function.

Let $\text{Opposite}$ denotes the length of the side opposite to the $30^{\circ}$ acute angle, and $\text{Adjacent}$ be the length of the side next to this $30^{\circ}$ acute angle.

$\displaystyle \begin{aligned}\text{Opposite} &= \text{Hypotenuse} \times \sin{30^{\circ}}\\ &=2\sqrt{3}\times \frac{1}{2} \\&= \sqrt{3}\end{aligned}$.

Similarly,

$\displaystyle \begin{aligned}\text{Adjacent} &= \text{Hypotenuse} \times \cos{30^{\circ}}\\ &=2\sqrt{3}\times \frac{\sqrt{3}}{2} \\&= 3\end{aligned}$.

The longer leg in this case is the one adjacent to the $30^{\circ}$ acute angle. The answer will be $3$.

There's a shortcut to the answer. Notice that $\sin{30^{\circ}} < \cos{30^{\circ}}$. The cosine of an acute angle is directly related to the adjacent leg. In other words, the leg adjacent to the $30^{\circ}$ angle will be the longer leg. There will be no need to find the length of the opposite leg.

Does this relationship $\sin{\theta} < \cos{\theta}$ holds for all acute angles? (That is, $0^{\circ} < \theta$?) It turns out that:

• $\sin{\theta} < \cos{\theta}$ if $0^{\circ} < \theta$;
• $\sin{\theta} > \cos{\theta}$ if $45^{\circ} < \theta$;
• $\sin{\theta} = \cos{\theta}$ if $\theta = 45^{\circ}$.