Question

Solve the triangle

a = 2, c = 1, B = 10$°$

Short answer

The solution is

$b=1.04,A=156.92°,C=13.08°$

Step-by-step solution

Step1. Given information

a = 2, c = 1, B = 10$°$

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. Calculation

We know

$$\begin{gathered} b^2=a^2+c^2-2ac.\cos B \\ =2^2+1^2-2.2.1.\cos 10° \\ =4+1-4.\left(.98\right) \\ =5-3.92 \\ =1.08 \\ b=\sqrt{1.08} \\ =1.04 \\ Again, \\ cosA=\frac{b^2+c^2-a^2}{2bc} \\ =\frac{{\left(1.04\right)}^2+1^2-2^2}{2.\left(1.04\right).1} \\ =\frac{1.08+1-4}{2.08} \\ =\frac{-1.92}{2.08} \\ =-.92 \\ A={\cos }^{-1}\left(-.92\right) \\ =156.92 \\ Againweknow \\ A+B+C=180° \\ 156.92°+10°+C=180° \\ C=180°-166.92 \\ C=13.08° \end{gathered}$$