How do you solve the triangle if b = 12, c=8, a=15?

Jul 9, 2018

$A \approx {95.08}^{\circ} , B \approx {52.83}^{\circ} , \mathmr{and} , C \approx {32.09}^{\circ}$.

Explanation:

Using the cosine formula, we get,

$\cos A = \frac{{b}^{2} + {c}^{2} - {a}^{2}}{2 b c}$,

$= \frac{{12}^{2} + {8}^{2} - {15}^{2}}{2 \cdot 12 \cdot 8}$,

$= \frac{144 + 64 - 225}{192}$,

$= - \frac{17}{192}$.

$\therefore A = \arccos \left(- \frac{17}{192}\right) \approx {95.08}^{\circ}$.

$B = \arccos \left(\frac{{c}^{2} + {a}^{2} - {b}^{2}}{2 c a}\right)$,

$= \arccos \left\{\frac{{8}^{2} + {15}^{2} - {12}^{2}}{2 \cdot 8 \cdot 15}\right\}$,

$= \arccos \left(\frac{145}{240}\right)$.

$\Rightarrow B \approx {52.83}^{\circ}$.

Finally, $C = \arccos \left(\frac{{a}^{2} + {b}^{2} - {c}^{2}}{2 a b}\right)$,

$= \arccos \left\{\frac{{15}^{2} + {12}^{2} - {8}^{2}}{2 \cdot 15 \cdot 12}\right\}$,

$= \arccos \left(\frac{305}{360}\right)$.

$\Rightarrow C \approx {32.09}^{\circ}$.

$\mathmr{and} , C = {180}^{\circ} - \left({95.08}^{\circ} + {52.83}^{\circ}\right) \approx {32.09}^{\circ}$.