Question

in $\triangle jkl$, $mangle j = 90^{circ}$, $mangle k = 30^{circ}$, and $mangle l = 60^{circ}$. which of the following statements about $\triangle jkl$ are true? check all that apply. a. $jk=\frac{sqrt{3}}{2}cdot kl$ b. $jk = sqrt{3}cdot jl$ c. $jk = 2cdot jl$ d. $kl = 2cdot jl$ e. $kl=sqrt{3}cdot jl$ f. $jl=\frac{sqrt{3}}{2}cdot kl$

Explanation:

Step1: Recall trigonometric ratios in right - triangle

In right - triangle $\triangle{JKL}$ with $\angle{J} = 90^{\circ}$, $\angle{K}=30^{\circ}$, and $\angle{L}=60^{\circ}$. Let the side opposite $\angle{K}$ be $JL$, the side opposite $\angle{L}$ be $JK$, and the hypotenuse be $KL$.
We know that $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$ and $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$.

Step2: Use sine and cosine of angles

$\sin K=\sin30^{\circ}=\frac{JL}{KL}=\frac{1}{2}$, so $KL = 2\cdot JL$.
$\cos K=\cos30^{\circ}=\frac{JK}{KL}=\frac{\sqrt{3}}{2}$, so $JK=\frac{\sqrt{3}}{2}KL$.
Also, $\tan K=\tan30^{\circ}=\frac{JL}{JK}=\frac{1}{\sqrt{3}}$, so $JK=\sqrt{3}\cdot JL$.

Answer:

A. $JK=\frac{\sqrt{3}}{2}\cdot KL$
B. $JK=\sqrt{3}\cdot JL$
D. $KL = 2\cdot JL$