Question

consider a triangle (abc) like the one below. suppose that (b = 40), (c = 12), and (b=63^{circ}). (the figure is not drawn to scale.) solve the triangle. 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\. (c=square^{circ},a = square^{circ},a=square)

Explanation:

Step1: Use the Law of Sines to find angle C

By the Law of Sines, $\frac{\sin C}{c}=\frac{\sin B}{b}$. Substituting $b = 40$, $c = 12$, and $B=63^{\circ}$, we get $\sin C=\frac{c\sin B}{b}=\frac{12\times\sin(63^{\circ})}{40}$.
$\sin C=\frac{12\times0.8910}{40}=\frac{10.692}{40}=0.2673$.
Then $C=\sin^{- 1}(0.2673)\approx15.5^{\circ}$.

Step2: Find angle A

Since the sum of angles in a triangle is $180^{\circ}$, $A = 180^{\circ}-B - C$. Substituting $B = 63^{\circ}$ and $C\approx15.5^{\circ}$, we have $A=180^{\circ}-63^{\circ}-15.5^{\circ}=101.5^{\circ}$.

Step3: Use the Law of Sines to find side a

Again, by the Law of Sines $\frac{a}{\sin A}=\frac{b}{\sin B}$. So $a=\frac{b\sin A}{\sin B}$. Substituting $b = 40$, $A = 101.5^{\circ}$, and $B = 63^{\circ}$, we get $a=\frac{40\times\sin(101.5^{\circ})}{\sin(63^{\circ})}=\frac{40\times0.9806}{0.8910}\approx44.0$.

Answer:

$C = 15.5^{\circ},A = 101.5^{\circ},a = 44.0$