Question

In the given problem solve the triangle using either the law of sines or law of cosines-

$A=10°,a=3,b=10$

Short answer

Required values of the triangle are$B=35.35°,C=134.65°,c=12.29$

Step-by-step solution

Step 1. Given information

In the given triangle,

$A=10°,a=3,b=10$

We know, Law of sines

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$ ------role="math" localid="1646587025370" $\left(1\right)$

Law of cosines,

$c^2=a^2+b^2-2ab\cos C$ ------$\left(2\right)$

Step 2. Calculation

$$\begin{gathered} ForB, \\ \frac{\sin A}{a}=\frac{\sin B}{b} \\ \sin B=\raisebox{1ex}{$\sin A\times b$}\!\left/ \!\raisebox{-1ex}{$a$}\right. \\ =\sin \left(10°\right)\times \raisebox{1ex}{$10$}\!\left/ \!\raisebox{-1ex}{$3$}\right. \\ =\raisebox{1ex}{$0.1736\times 10$}\!\left/ \!\raisebox{-1ex}{$3$}\right. \\ =\raisebox{1ex}{$1.736$}\!\left/ \!\raisebox{-1ex}{$3$}\right. \\ =0.5786 \\ B={\sin }^{-1}\left(0.5786\right) \\ =35.35° \\ ForC, \\ C=180°-A-B \\ =180°-10°-35.35° \\ =134.65° \\ Forc, \\ {c}^2=a^2+b^2-2ab\cos \left(C\right) \\ =3^2+{10}^2-2\times 3\times 10\times \cos \left(134.65°\right) \\ =9+100-60\times \left(-0.7027\right) \\ =151.162 \\ c=\sqrt{151.162}=12.29 \end{gathered}$$