Question

in the 30 - 60 - 90 triangle below, side s has a length of __ and the hypotenuse has a length of __. a. 1; √3 b. 1; √2 c. √2; 2 d. 1; 1.7 e. 2; 1.4 f. √3; 2

Explanation:

Step1: Recall 30 - 60 - 90 triangle ratio

In a 30 - 60 - 90 triangle, if the side opposite the 30° angle is $x$, the side opposite the 60° angle is $x\sqrt{3}$ and the hypotenuse is $2x$. Here the side opposite the 60° angle is given as 1. Let the side opposite the 30° angle be $s$ and the hypotenuse be $h$.
We know that $\tan60^{\circ}=\frac{s}{1}$, and $\sin60^{\circ}=\frac{1}{h}$.
Since $\tan60^{\circ}=\sqrt{3}=\frac{s}{1}$, then $s = \sqrt{3}$.

Step2: Find the hypotenuse

Since $\sin60^{\circ}=\frac{\sqrt{3}}{2}=\frac{1}{h}$, we can cross - multiply to get $h\times\sqrt{3}=2$, so $h = \frac{2}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{3}$ (using the ratio approach: if the side opposite 60° is 1, and the ratio of side opposite 60° to hypotenuse is $\frac{\sqrt{3}}{2}$, then if $a = 1$ and $\frac{a}{h}=\frac{\sqrt{3}}{2}$, $h=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}$). Also, using the Pythagorean theorem $h^{2}=s^{2}+1^{2}$, substituting $s = \sqrt{3}$, we have $h^{2}=3 + 1=4$, so $h = 2$.

Answer:

F. $\sqrt{3};2$