Question

what is the approximate length of $overline{kl}$? use the law of sines to find the answer. 1.8 units 2.0 units 3.2 units 3.7 units law of sines: $\frac{sin(a)}{a}=\frac{sin(b)}{b}=\frac{sin(c)}{c}$

Explanation:

Step1: Find angle J and K

The sum of angles in a triangle is 180°. Let $\angle L = 105^{\circ}$, assume $\angle J$ and $\angle K$ are the other two angles. Since the triangle is isosceles - like (not given explicitly but we can use the law of sines), we first find the other angles. $\angle J+\angle K = 180 - 105=75^{\circ}$. Let's assume the triangle is non - isosceles and we use the law of sines $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$. Let $JL = 2.7$, $JK = 4.7$, and we want to find $KL$. By the law of sines $\frac{KL}{\sin J}=\frac{JK}{\sin L}$. First, we find $\sin J$. We know that $\frac{2.7}{\sin K}=\frac{4.7}{\sin105^{\circ}}$, so $\sin K=\frac{2.7\sin105^{\circ}}{4.7}$. $\sin105^{\circ}=\sin(60^{\circ} + 45^{\circ})=\sin60^{\circ}\cos45^{\circ}+\cos60^{\circ}\sin45^{\circ}=\frac{\sqrt{3}}{2}\times\frac{\sqrt{2}}{2}+\frac{1}{2}\times\frac{\sqrt{2}}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}\approx0.9659$. Then $\sin K=\frac{2.7\times0.9659}{4.7}\approx0.55$. So $K\approx33.4^{\circ}$, and $J = 180-(105 + 33.4)=41.6^{\circ}$.

Step2: Use the law of sines to find KL

Now, using the law of sines $\frac{KL}{\sin J}=\frac{JK}{\sin L}$. Substitute $JK = 4.7$, $\sin J=\sin41.6^{\circ}\approx0.66$, $\sin L=\sin105^{\circ}\approx0.9659$. Then $KL=\frac{4.7\times\sin41.6^{\circ}}{\sin105^{\circ}}=\frac{4.7\times0.66}{0.9659}\approx3.2$.

Answer:

3.2 units