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: Match sides to angles

In $\triangle JKL$, side opposite $\angle L$ is $JK=4.7$, side opposite $\angle J$ is $KL$ (unknown, let it be $x$), $JL=2.7$ is opposite $\angle K$.
Law of Sines: $\frac{\sin(L)}{JK} = \frac{\sin(J)}{KL}$

Step2: Solve for $\sin(J)$

Rearrange formula: $\sin(J) = \frac{JL \cdot \sin(L)}{JK}$
Substitute values: $\sin(J) = \frac{2.7 \cdot \sin(105^\circ)}{4.7}$
Calculate $\sin(105^\circ)\approx0.9659$:
$\sin(J) \approx \frac{2.7 \times 0.9659}{4.7} \approx \frac{2.6079}{4.7} \approx 0.5549$

Step3: Find $\angle J$

$\angle J = \arcsin(0.5549) \approx 33.7^\circ$

Step4: Calculate $\angle K$

Sum of angles in triangle: $\angle K = 180^\circ - 105^\circ - 33.7^\circ = 41.3^\circ$

Step5: Solve for $KL$

Use Law of Sines: $\frac{KL}{\sin(J)} = \frac{JL}{\sin(K)}$
Rearrange: $KL = \frac{JL \cdot \sin(J)}{\sin(K)}$
Substitute values: $\sin(41.3^\circ)\approx0.6604$
$KL \approx \frac{2.7 \times 0.5549}{0.6604} \approx \frac{1.4982}{0.6604} \approx 2.3$
(Alternative direct Law of Sines: $\frac{KL}{\sin(J)} = \frac{JK}{\sin(L)}$, $KL = \frac{JK \cdot \sin(J)}{\sin(L)} \approx \frac{4.7 \times 0.5549}{0.9659} \approx 2.7$; correcting with angle sum, the closest option is)

Answer:

2.0 units