Question

In the given problem solve the triangle using the law of sines : $B=10°,C=100°,b=2$

Short answer

Required values of the triangle are$c=11.3,a=10.8,A=70°$

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

$$\begin{gathered} Given, \\ B=10°,C=100°,b=2 \\ Forc, \\ \Rightarrow \raisebox{1ex}{$\sin \left(B\right)$}\!\left/ \!\raisebox{-1ex}{$b$}\right.=\raisebox{1ex}{$\sin \left(C\right)$}\!\left/ \!\raisebox{-1ex}{$c$}\right. \\ \Rightarrow c=\raisebox{1ex}{$\sin \left(C\right)\times b$}\!\left/ \!\raisebox{-1ex}{$\sin \left(B\right)$}\right. \\ =\raisebox{1ex}{$\sin \left(100°\right)\times 2$}\!\left/ \!\raisebox{-1ex}{$\sin \left(10°\right)$}\right. \\ =\raisebox{1ex}{$0.9848\times 2$}\!\left/ \!\raisebox{-1ex}{$0.1736$}\right. \\ =\raisebox{1ex}{$1.9696$}\!\left/ \!\raisebox{-1ex}{$0.1736$}\right. \\ =11.345\approx 11.3 \\ ForA, \\ \Rightarrow A=180°-B-C \\ =180°-10°-100° \\ =70° \\ Fora, \\ \Rightarrow \raisebox{1ex}{$\sin \left(A\right)$}\!\left/ \!\raisebox{-1ex}{$a$}\right.=\raisebox{1ex}{$\sin \left(B\right)$}\!\left/ \!\raisebox{-1ex}{$b$}\right. \\ \Rightarrow a=\raisebox{1ex}{$\sin \left(A\right)\times b$}\!\left/ \!\raisebox{-1ex}{$\sin \left(B\right)$}\right. \\ =\raisebox{1ex}{$\sin \left(70°\right)\times 2$}\!\left/ \!\raisebox{-1ex}{$\sin \left(10°\right)$}\right. \\ =\raisebox{1ex}{$0.9396\times 2$}\!\left/ \!\raisebox{-1ex}{$0.1736$}\right. \\ =\raisebox{1ex}{$1.8792$}\!\left/ \!\raisebox{-1ex}{$0.1736$}\right. \\ =10.824\approx 10.8 \end{gathered}$$