Find the longer leg of the triangle.A. 3B. [tex]\sqrt{3}[/tex]C. 9D. [tex]\sqrt{6}[/tex]

Accepted Solution

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 [tex]\text{Opposite}[/tex] denotes the length of the side opposite to the [tex]30^{\circ}[/tex] acute angle, and [tex]\text{Adjacent}[/tex] be the length of the side next to this [tex]30^{\circ}[/tex] acute angle.[tex]\displaystyle \begin{aligned}\text{Opposite} &= \text{Hypotenuse} \times \sin{30^{\circ}}\\ &=2\sqrt{3}\times \frac{1}{2} \\&= \sqrt{3}\end{aligned}[/tex].Similarly, [tex]\displaystyle \begin{aligned}\text{Adjacent} &= \text{Hypotenuse} \times \cos{30^{\circ}}\\ &=2\sqrt{3}\times \frac{\sqrt{3}}{2} \\&= 3\end{aligned}[/tex].The longer leg in this case is the one adjacent to the [tex]30^{\circ}[/tex] acute angle. The answer will be [tex]3[/tex].There's a shortcut to the answer. Notice that [tex]\sin{30^{\circ}} < \cos{30^{\circ}}[/tex]. The cosine of an acute angle is directly related to the adjacent leg. In other words, the leg adjacent to the [tex]30^{\circ}[/tex] angle will be the longer leg. There will be no need to find the length of the opposite leg. Does this relationship [tex]\sin{\theta} < \cos{\theta}[/tex] holds for all acute angles? (That is, [tex]0^{\circ} < \theta <90^{\circ}[/tex]?) It turns out that:[tex]\sin{\theta} < \cos{\theta}[/tex] if [tex]0^{\circ} < \theta <45^{\circ}[/tex];[tex]\sin{\theta} > \cos{\theta}[/tex] if [tex]45^{\circ} < \theta <90^{\circ}[/tex];[tex]\sin{\theta} = \cos{\theta}[/tex] if [tex]\theta = 45^{\circ}[/tex].