In triangle ABC, a=15, b=14, c=10. Find m&lt;B

Question

In triangle ABC, a=15, b=14, c=10. Find m&amp;lt;B

yunitsiL · Accepted Answer

The given case is SSS (three sides) so use the Law of Cosines: $b^{2} = a^{2} + c^{2} - 2 a o s B$
Solve for B.
$2 a o s B = a^{2} + c^{2} - b^{2}$
$\cos B = \frac{a^{2} + c^{2} - b^{2}}{2} a c$
$m < B = (\cos^{- 1}) (\frac{a^{2} + c^{2} - b^{2}}{2 a c})$
$m < B = (\cos^{- 1}) (\frac{15^{2} + 10^{2} - 14^{2}}{2 (15) (10)})$
$m < B = \cos^{- 1} (\frac{129}{300})$
Use a calculator in DEGREE mode: $m ∠ B \approx 64.5 °$

In triangle ABC, a=15, b=14, c=10. Find m<B

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