Question

In the given problem solve the triangle using the law of sines : $B=20°,C=70°,a=1$

Short answer

Required values of the triangle are$A=90°,b=0.3,c=0.9$

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

$$\begin{gathered} Given, \\ B=20°,C=70°,a=1 \\ ForA, \\ \Rightarrow A=180°-B-C \\ =180°-20°-70° \\ =90° \\ Forc, \\ \Rightarrow \raisebox{1ex}{$\sin \left(A\right)$}\!\left/ \!\raisebox{-1ex}{$a$}\right.=\raisebox{1ex}{$\sin \left(C\right)$}\!\left/ \!\raisebox{-1ex}{$c$}\right. \\ \Rightarrow c=\raisebox{1ex}{$\sin \left(C\right)\times a$}\!\left/ \!\raisebox{-1ex}{$\sin \left(A\right)$}\right. \\ =\raisebox{1ex}{$\sin \left(70°\right)\times 1$}\!\left/ \!\raisebox{-1ex}{$\sin \left(90°\right)$}\right. \\ =\raisebox{1ex}{$0.9396\times 1$}\!\left/ \!\raisebox{-1ex}{$1$}\right. \\ =0.9396\approx 0.9 \\ Forb, \\ \Rightarrow \raisebox{1ex}{$\sin \left(A\right)$}\!\left/ \!\raisebox{-1ex}{$a$}\right.=\raisebox{1ex}{$\sin \left(B\right)$}\!\left/ \!\raisebox{-1ex}{$b$}\right. \\ \Rightarrow b=\raisebox{1ex}{$\sin \left(B\right)\times a$}\!\left/ \!\raisebox{-1ex}{$\sin \left(A\right)$}\right. \\ =\raisebox{1ex}{$\sin \left(20°\right)\times 1$}\!\left/ \!\raisebox{-1ex}{$\sin \left(90°\right)$}\right. \\ =\raisebox{1ex}{$0.342\times 1$}\!\left/ \!\raisebox{-1ex}{$1$}\right. \\ =0.342\approx 0.3 \end{gathered}$$