Maths Question:
$\text{Where possible solve the triangle }c=2.718,\text{ }b=3.142,\text{ }C={{56}^{\circ }}.18'$
Maths Solution:
$\begin{align} & \text{Using sine rule }\frac{b}{\sin B}=\frac{c}{\sin C} \\ & \sin B=\frac{b\sin C}{c}=\frac{3.142\times \sin {{56}^{\circ }}18'}{2.718}=0.9617 \\ & B={{\sin }^{-1}}0.9617={{74}^{\circ }}6' \\ & A={{180}^{\circ }}-(B+C)=180-({{74}^{\circ }}{{6}^{'}}+{{56}^{\circ }}18') \\ & A={{49}^{\circ }}36' \\\end{align}$
University mathstopic: