Question

what is the length of line segment rs? use the law of sines to find the answer. round to the nearest tenth. law of sines: $\frac{sin(a)}{a}=\frac{sin(b)}{b}=\frac{sin(c)}{c}$ 2.2 units 2.4 units 3.0 units 3.3 units

Explanation:

Step1: Identify known values

Let $QR = 2.4$, $\angle R=80^{\circ}$, $QS = 3.1$. We want to find $RS$. By the law of sines $\frac{\sin(A)}{a}=\frac{\sin(B)}{b}=\frac{\sin(C)}{c}$, in $\triangle QRS$, $\frac{\sin R}{QS}=\frac{\sin S}{QR}$. First, find $\sin S$.
$\sin S=\frac{QR\times\sin R}{QS}$
Substitute $QR = 2.4$, $\angle R = 80^{\circ}$ ($\sin80^{\circ}\approx0.9848$), $QS = 3.1$:
$\sin S=\frac{2.4\times0.9848}{3.1}\approx\frac{2.3635}{3.1}\approx0.7624$

Step2: Find angle $S$

$S=\sin^{- 1}(0.7624)\approx49.6^{\circ}$

Step3: Find angle $Q$

Since the sum of angles in a triangle is $180^{\circ}$, $\angle Q=180^{\circ}-\angle R-\angle S$.
$\angle Q = 180^{\circ}-80^{\circ}-49.6^{\circ}=50.4^{\circ}$

Step4: Use law of sines to find $RS$

Now use $\frac{\sin Q}{RS}=\frac{\sin R}{QS}$, so $RS=\frac{QS\times\sin Q}{\sin R}$
Substitute $QS = 3.1$, $\sin Q\approx0.7704$ (for $Q = 50.4^{\circ}$), $\sin R\approx0.9848$ (for $R = 80^{\circ}$)
$RS=\frac{3.1\times0.7704}{0.9848}\approx2.4$

Answer:

B. 2.4 units