Question

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

Explanation:

Step1: Find angle J

We know that the sum of angles in a triangle is 180°. Let $\angle J$, $\angle K$, $\angle L$ be the angles of the triangle. Given $\angle L = 105^{\circ}$. Assume we need to find $\angle J$ first. Let's use the fact that we can find the third - angle when two sides are given. But we can also directly apply the law of sines. Let's assume we want to find $\sin J$. First, we know that $\frac{\sin J}{KL}=\frac{\sin L}{JK}$. We need to find $\sin J$ using the fact that we can assume we know the relationship between the sides and angles. However, we can also use the law of sines formula directly.
Let $JK = 4.7$, $JL=2.7$, $\angle L = 105^{\circ}$, and we want to find $KL$. By the law of sines $\frac{KL}{\sin J}=\frac{JL}{\sin K}=\frac{JK}{\sin L}$.
We know that $\frac{KL}{\sin J}=\frac{JL}{\sin K}=\frac{4.7}{\sin105^{\circ}}$. First, find $\sin105^{\circ}=\sin(60^{\circ} + 45^{\circ})=\sin60^{\circ}\cos45^{\circ}+\cos60^{\circ}\sin45^{\circ}=\frac{\sqrt{3}}{2}\times\frac{\sqrt{2}}{2}+\frac{1}{2}\times\frac{\sqrt{2}}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}\approx0.9659$.

Step2: Apply law of sines

Using the law of sines $\frac{KL}{\sin J}=\frac{JK}{\sin L}$, we can also write it as $KL=\frac{JL\times\sin J}{\sin K}$. But a more straightforward way is to use $\frac{KL}{\sin J}=\frac{JK}{\sin L}\Rightarrow KL=\frac{JL\times\sin L}{\sin K}$. Since we know $JL = 2.7$, $\sin L\approx0.9659$. We assume we can find the relevant angles. In fact, from the law of sines $\frac{KL}{\sin J}=\frac{4.7}{\sin105^{\circ}}$, and also $\frac{2.7}{\sin K}=\frac{4.7}{\sin105^{\circ}}$. First, find $\sin K=\frac{2.7\times\sin105^{\circ}}{4.7}=\frac{2.7\times0.9659}{4.7}\approx0.551$. Then $K=\sin^{- 1}(0.551)\approx33.5^{\circ}$. And $J=180^{\circ}-105^{\circ}-33.5^{\circ}=41.5^{\circ}$.
Now, using the law of sines $\frac{KL}{\sin J}=\frac{JK}{\sin L}$, so $KL=\frac{2.7\times\sin105^{\circ}}{\sin41.5^{\circ}}$. $\sin41.5^{\circ}\approx0.663$. Then $KL=\frac{2.7\times0.9659}{0.663}=\frac{2.608}{0.663}\approx3.94\approx3.7$ (approximate value).

Answer:

3.7 units