How do you solve the triangle given m∠B = 117°, a = 16, b = 38?

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Answer 1 · 2016-10-25T16:34:42

Please see the explanation for the process of "How to".

Using the Law of Sines , we can find the measure of $∠ A$ $/_A$

$∠ A = {sin}^{- 1} ((\frac{a}{b}) sin (∠ B))$ $/_A = sin^-1((a/b)sin(/_B))$

$∠ A = {sin}^{- 1} ((\frac{16}{38}) sin (117^{\circ}))$ $/_A = sin^-1((16/38)sin(117^@))$

$∠ A \approx 22^{\circ}$ $/_A ~~ 22^@$

To find the measure of $∠ C$ $/_C$ , use the fact that the sum of the angles of a triangle is 180°.

$∠ C = 180^{\circ} - 117^{\circ} - 22^{\circ}$ $/_C = 180^@ - 117^@ - 22^@$

$∠ C = 41^{\circ}$ $/_C = 41^@$

Use the Law of Sines to find the length of side c:

$c = a \frac{sin (∠ C)}{sin (∠ A)}$ $c = asin(/_C)/sin(/_A)$

$c = 16 \frac{sin (41^{\circ})}{sin (22^{\circ})}$ $c = 16sin(41^@)/sin(22^@)$

$c \approx 28.0$ $c ~~28.0$