Question

In a triangle $ABC$, angle B has a size of ${50}^{\circ }$, angle A has a size of ${32}^{\circ }$ and the length of side $BC$ is $150$ units. The length of side $AB$ is (approximately):

• A. $232$
• B. $250$
• C. $260$
• D. $270$
• E. $280$

Answer 1

The correct option is E $280$

In triangle $ABC$,

$\angle A+\angle B+\angle C={180}^{\circ }$

${32}^{\circ }+{50}^{\circ }+\angle C={180}^{\circ }$

$\angle C={180}^{\circ }-{32}^{\circ }-{50}^{\circ }={98}^{\circ }$

We have from Sin law,

$\frac{\sin A}{BC}=\frac{\sin B}{AC}=\frac{\sin C}{AB}$

Consider,

$\frac{\sin A}{BC}=\frac{\sin C}{AB}$

$AB=\frac{\sin C}{\sin A}\left(BC\right)$

$AB=\frac{\sin {98}^{\circ }}{\sin {32}^{\circ }}\left(150\right)$

$AB=\frac{0.99}{0.5299}\left(150\right)$

$AB=280.30units$