Question

In the given problem solve the triangle using the law of sines :

$A=40°,B=20°,a=2.$

Short answer

Required values of the triangle are$C=120°,b=1.1,c=2.7$

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

Given,

$$\begin{gathered} A=40°,B=20°,a=2 \\ AccordingtoLawofSines, \\ Forb, \\ \Rightarrow \raisebox{1ex}{$\sin \left(A\right)$}\!\left/ \!\raisebox{-1ex}{$a$}\right.=\raisebox{1ex}{$\sin \left(B\right)$}\!\left/ \!\raisebox{-1ex}{$b$}\right. \\ \Rightarrow b=\raisebox{1ex}{$\sin \left(B\right)\times a$}\!\left/ \!\raisebox{-1ex}{$\sin \left(A\right)$}\right. \\ =\raisebox{1ex}{$\sin \left(20°\right)\times 2$}\!\left/ \!\raisebox{-1ex}{$\sin \left(40°\right)$}\right. \\ =\raisebox{1ex}{$0.342\times 2$}\!\left/ \!\raisebox{-1ex}{$0.6427$}\right. \\ =\raisebox{1ex}{$0.684$}\!\left/ \!\raisebox{-1ex}{$0.6427$}\right. \\ =1.064\approx 1.1 \\ ForC, \\ C=180°-A-B \\ =180°-40°-20° \\ =120° \\ Forc, \\ \Rightarrow \raisebox{1ex}{$\sin \left(A\right)$}\!\left/ \!\raisebox{-1ex}{$a$}\right.=\raisebox{1ex}{$\sin \left(C\right)$}\!\left/ \!\raisebox{-1ex}{$c$}\right. \\ \Rightarrow c=\raisebox{1ex}{$\sin \left(C\right)\times a$}\!\left/ \!\raisebox{-1ex}{$\sin \left(A\right)$}\right. \\ =\raisebox{1ex}{$\sin \left(120°\right)\times 2$}\!\left/ \!\raisebox{-1ex}{$\sin \left(40°\right)$}\right. \\ =\raisebox{1ex}{$0.866\times 2$}\!\left/ \!\raisebox{-1ex}{$0.6427$}\right. \\ =\raisebox{1ex}{$1.732$}\!\left/ \!\raisebox{-1ex}{$0.6427$}\right. \\ =2.694\approx 2.7 \end{gathered}$$