Draw the triangle ABC.
Determine the length of side AC.
[tex]\begin{gathered} (AC)^2=(AB)^2+(BC)^2-2\cdot AB\cdot BC\cdot\cos B \\ =(12)^2+(18)^2-2\cdot12\cdot18\cdot\cos 75 \\ =356.190 \\ AC=\sqrt[]{356.19} \\ =18.87 \\ \approx18.9 \end{gathered}[/tex]So side AC is equal to 18.9 m.
Determine the measure of angle A.
[tex]\begin{gathered} \frac{AC}{\sin B}=\frac{BC}{\sin A} \\ \frac{18.9}{\sin75}=\frac{18}{\sin A} \\ \sin A=\frac{18}{18.9}\cdot\sin 75 \\ A=\sin ^{-1}(0.9199) \\ =66.9 \end{gathered}[/tex]So mesure of angle A is 66.9 degree.
