Question

In the given problem solve the triangle using the law of sines :$A=110°,C=30°,c=3$

Short answer

Required values of the triangle are$a=5.6,b=3.9,B=40°$

Step-by-step solution

Step 1. Given information

For a triangle with sides a, b, c and opposite angles A, B, C, respectively, Law of sines is given as:

$\raisebox{1ex}{$\sin \left(A\right)$}\!\left/ \!\raisebox{-1ex}{$a$}\right.=\raisebox{1ex}{$\sin \left(B\right)$}\!\left/ \!\raisebox{-1ex}{$b$}\right.=\raisebox{1ex}{$\sin \left(C\right)$}\!\left/ \!\raisebox{-1ex}{$c$}\right.$

Step 2. Calculation

$$\begin{gathered} Given, \\ A=110°,C=30°,c=3 \\ AccordingtoLawofSines, \\ Fora, \\ \Rightarrow \raisebox{1ex}{$\sin \left(A\right)$}\!\left/ \!\raisebox{-1ex}{$a$}\right.=\raisebox{1ex}{$\sin \left(C\right)$}\!\left/ \!\raisebox{-1ex}{$c$}\right. \\ \Rightarrow a=\raisebox{1ex}{$\sin \left(A\right)\times c$}\!\left/ \!\raisebox{-1ex}{$\sin \left(C\right)$}\right. \\ =\raisebox{1ex}{$\sin \left(110°\right)\times 3$}\!\left/ \!\raisebox{-1ex}{$\sin \left(30°\right)$}\right. \\ =\raisebox{1ex}{$0.9396\times 3$}\!\left/ \!\raisebox{-1ex}{$0.5$}\right. \\ =\raisebox{1ex}{$2.8188$}\!\left/ \!\raisebox{-1ex}{$0.5$}\right. \\ =5.6376\approx 5.6 \\ ForB, \\ B=180°-A-C \\ =180°-110°-30° \\ =40° \\ Forb, \\ \Rightarrow \raisebox{1ex}{$\sin \left(B\right)$}\!\left/ \!\raisebox{-1ex}{$b$}\right.=\raisebox{1ex}{$\sin \left(C\right)$}\!\left/ \!\raisebox{-1ex}{$c$}\right. \\ \Rightarrow b=\raisebox{1ex}{$\sin \left(B\right)\times c$}\!\left/ \!\raisebox{-1ex}{$\sin \left(C\right)$}\right. \\ =\raisebox{1ex}{$\sin \left(40°\right)\times 3$}\!\left/ \!\raisebox{-1ex}{$\sin \left(30°\right)$}\right. \\ =\raisebox{1ex}{$0.6427\times 3$}\!\left/ \!\raisebox{-1ex}{$0.5$}\right. \\ =\raisebox{1ex}{$1.9281$}\!\left/ \!\raisebox{-1ex}{$0.5$}\right. \\ =3.8562\approx 3.9 \end{gathered}$$