Question

law of sines: $\frac{sin(a)}{a}=\frac{sin(b)}{b}=\frac{sin(c)}{c}$ in $\triangle abc$, $c = 5.4$, $a = 3.3$, and $mangle a=20^{circ}$. what are the possible approximate lengths of b? use the law of sines to find the answer. 2.0 units and 4.6 units 2.1 units and 8.7 units 2.3 units and 7.8 units

Explanation:

Step1: Use the law of sines to find $\sin C$.

$\frac{\sin A}{a}=\frac{\sin C}{c}$, so $\sin C=\frac{c\sin A}{a}$. Substitute $a = 3.3$, $c = 5.4$, and $A = 20^{\circ}$ (so $\sin A=\sin(20^{\circ})\approx0.342$). Then $\sin C=\frac{5.4\times0.342}{3.3}=\frac{1.8468}{3.3}\approx0.5596$.

Step2: Find the two possible values of $C$.

$C_1=\sin^{- 1}(0.5596)\approx34^{\circ}$ and $C_2 = 180^{\circ}-34^{\circ}=146^{\circ}$.

Step3: Find the two possible values of $B$.

For $C_1 = 34^{\circ}$, $B_1=180^{\circ}-(20^{\circ}+34^{\circ}) = 126^{\circ}$. For $C_2=146^{\circ}$, $B_2=180^{\circ}-(20^{\circ}+146^{\circ})=14^{\circ}$.

Step4: Use the law of sines to find the two possible values of $b$.

Using $\frac{\sin A}{a}=\frac{\sin B}{b}$, for $B_1 = 126^{\circ}$, $b_1=\frac{a\sin B_1}{\sin A}=\frac{3.3\times\sin(126^{\circ})}{0.342}=\frac{3.3\times0.809}{0.342}\approx7.8$. For $B_2 = 14^{\circ}$, $b_2=\frac{a\sin B_2}{\sin A}=\frac{3.3\times\sin(14^{\circ})}{0.342}=\frac{3.3\times0.242}{0.342}\approx2.3$.

Answer:

2.3 units and 7.8 units