Question

Solve the triangle

a = 3, b = 4, C = 40$°$

Short answer

The solution is

$c=2.55,A=48.7°,B=91.3°$

Step-by-step solution

Step1. Given information

$a=3,b=4,C=40°$

Step2. Formula used

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

Step3. Calculation

$$\begin{gathered} Weknow \\ {c}^2=a^2+b^2-2ab.\cos C \\ =3^2+4^2-2.3.4.\cos 40° \\ =9+16-24.\left(.77\right) \\ =25-18.48 \\ =6.52 \\ c=\sqrt{6.52} \\ =2.55 \\ Again, \\ \cos A=\frac{b^2+c^2-a^2}{2bc} \\ =\frac{4^2+{\left(2.55\right)}^2-3^2}{2.4.\left(2.55\right)} \\ =\frac{16+6.52-9}{20.4} \\ =\frac{13.52}{20.4} \\ =.66 \\ A={\cos }^{-1}\left(.66\right) \\ =48.7° \\ weknow, \\ A+B+C=180° \\ 48.7°+B+40°=180° \\ B=180°-88.7° \\ =91.3° \end{gathered}$$