Question

consider a triangle (abc) like the one below. suppose that (b = 27), (c = 31), and (b=36^{circ}). (the figure is not drawn to scale.) carry your intermediate computations to at least four - decimal places, and round your answers to the nearest tenth. if no such triangle exists, enter
o solution.\ if there is more than one solution, use the button labeled \or\.

Explanation:

Step1: Use the Law of Sines

By the Law of Sines, $\frac{\sin B}{b}=\frac{\sin C}{c}$. Substitute $b = 27$, $c = 31$, and $B=36^{\circ}$ into the formula: $\sin C=\frac{c\sin B}{b}=\frac{31\sin36^{\circ}}{27}$.
$\sin C=\frac{31\times0.587785}{27}\approx\frac{18.221335}{27}\approx0.6749$.

Step2: Find angle C

Since $\sin C\approx0.6749$, then $C=\sin^{- 1}(0.6749)$ or $C = 180^{\circ}-\sin^{-1}(0.6749)$.
$C_1=\sin^{-1}(0.6749)\approx42.4^{\circ}$ and $C_2 = 180^{\circ}-42.4^{\circ}=137.6^{\circ}$.
But when $C = 137.6^{\circ}$, $B + C=36^{\circ}+137.6^{\circ}=173.6^{\circ}<180^{\circ}$.

Step3: Find angle A

For $C = 42.4^{\circ}$, $A=180^{\circ}-B - C=180^{\circ}-36^{\circ}-42.4^{\circ}=101.6^{\circ}$.
For $C = 137.6^{\circ}$, $A=180^{\circ}-B - C=180^{\circ}-36^{\circ}-137.6^{\circ}=6.4^{\circ}$.

Step4: Find side a

Using the Law of Sines $\frac{\sin A}{a}=\frac{\sin B}{b}$.
When $A = 101.6^{\circ}$, $a=\frac{b\sin A}{\sin B}=\frac{27\sin101.6^{\circ}}{\sin36^{\circ}}=\frac{27\times0.9806}{0.5878}\approx45.2$.
When $A = 6.4^{\circ}$, $a=\frac{b\sin A}{\sin B}=\frac{27\sin6.4^{\circ}}{\sin36^{\circ}}=\frac{27\times0.1119}{0.5878}\approx5.2$.

So the two - solution cases are:
Case 1: $C = 42.4^{\circ}$, $A = 101.6^{\circ}$, $a = 45.2$
Case 2: $C = 137.6^{\circ}$, $A = 6.4^{\circ}$, $a = 5.2$

Answer:

$C = 42.4^{\circ},A = 101.6^{\circ},a = 45.2$ or $C = 137.6^{\circ},A = 6.4^{\circ},a = 5.2$