Question

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

$A=40°,B=40°,c=2$

Short answer

Required values of the triangle are$C=100°,a=1.3,b=1.3$

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

$$\begin{gathered} Given, \\ A=40°,B=40°,c=2 \\ ForC, \\ C=180°-A-B \\ =180°-40°-40° \\ =100° \\ AccordingtoLawofSines, \\ Forb, \\ \Rightarrow \raisebox{1ex}{$\sin \left(C\right)$}\!\left/ \!\raisebox{-1ex}{$c$}\right.=\raisebox{1ex}{$\sin \left(B\right)$}\!\left/ \!\raisebox{-1ex}{$b$}\right. \\ \Rightarrow b=\raisebox{1ex}{$\sin \left(B\right)\times c$}\!\left/ \!\raisebox{-1ex}{$\sin \left(C\right)$}\right. \\ =\raisebox{1ex}{$\sin \left(40°\right)\times 2$}\!\left/ \!\raisebox{-1ex}{$\sin \left(100°\right)$}\right. \\ =\raisebox{1ex}{$0.6427\times 2$}\!\left/ \!\raisebox{-1ex}{$0.9848$}\right. \\ =\raisebox{1ex}{$1.2854$}\!\left/ \!\raisebox{-1ex}{$0.9848$}\right. \\ =1.305\approx 1.3 \\ Fora, \\ \Rightarrow \raisebox{1ex}{$\sin \left(A\right)$}\!\left/ \!\raisebox{-1ex}{$a$}\right.=\raisebox{1ex}{$\sin \left(B\right)$}\!\left/ \!\raisebox{-1ex}{$b$}\right. \\ \Rightarrow a=\raisebox{1ex}{$\sin \left(A\right)\times b$}\!\left/ \!\raisebox{-1ex}{$\sin \left(B\right)$}\right. \\ =\raisebox{1ex}{$\sin \left(40°\right)\times 1.3$}\!\left/ \!\raisebox{-1ex}{$\sin \left(40°\right)$}\right. \\ =1.3 \end{gathered}$$