Question

Solve the triangle

Short answer

The solution is$b=3.19,A=11.48°,C=148.52°$

Step-by-step solution

Step1. Given information

Step2. Formula used

$$\begin{gathered} b^2=a^2+c^2-2ac.\cos B, \\ \cos A=\frac{b^2+c^2-a^2}{2bc} \end{gathered}$$

Step3. Calulation

We know,

$$\begin{gathered} b^2=a^2+c^2-2ac.\cos B \\ =2^2+5^2-2.2.5.\cos 20° \\ =4+25-20\times .939 \\ =29-18.78 \\ =10.22 \\ b=\sqrt{10.22} \\ =3.19 \\ Again, \\ \cos A=\frac{b^2+c^2-a^2}{2bc} \\ =\frac{{\left(3.19\right)}^2+5^2-2^2}{2.\left(3.19\right).5} \\ =\frac{10.22+25-4}{31.9} \\ =\frac{31.22}{31.9} \\ =.98 \\ A={\cos }^{-1}(.98) \\ =11.48° \\ Weknow, \\ A+B+C=180° \\ 11.48°+20°+C=180° \\ C=180°-31.48° \\ =148.52° \end{gathered}$$