Question

Select the correct answer.In triangle ABC, AB = 12, BC = 18, and m B = 75° what are the approximate length of side AC and measure of A?O AAC = 18.9;m SA = 66.99OB.OC.AC = 20.3 m A = 34.8°AC = 18.9: m A = 37.8°AC = 20.31 m A = 58.9°ODResetNext

Answer 1

Draw the triangle ABC.

Determine the length of side AC.

$\begin{gathered} (AC)^2=(AB)^2+(BC)^2-2\cdot AB\cdot BC\cdot\cos B \\ =(12)^2+(18)^2-2\cdot12\cdot18\cdot\cos 75 \\ =356.190 \\ AC=\sqrt[]{356.19} \\ =18.87 \\ \approx18.9 \end{gathered}$

So side AC is equal to 18.9 m.

Determine the measure of angle A.

$\begin{gathered} \frac{AC}{\sin B}=\frac{BC}{\sin A} \\ \frac{18.9}{\sin75}=\frac{18}{\sin A} \\ \sin A=\frac{18}{18.9}\cdot\sin 75 \\ A=\sin ^{-1}(0.9199) \\ =66.9 \end{gathered}$

So mesure of angle A is 66.9 degree.