How do you solve the triangle given m∠C = 13°, m∠A = 22°, c = 9?

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How do you solve the triangle given m∠C = 13°, m∠A = 22°, c = 9?

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Answer 1 · 2016-09-18T14:46:10

Start by drawing a diagram.

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We know an angle opposite a side, so we will be using The Law of Sines to solve this problem. Note that this is not an ambiguous case, since we know a side and two angles and not two sides and one angle.

$\frac{sin A}{a} = \frac{sin B}{b} = \frac{sin C}{c}$ $sinA/a = sinB/b = sinC/c$

$\frac{sin A}{a} = \frac{sin C}{c}$ $sinA/a = sinC/c$

$(sin22˚)/a = (sin13˚)/9$

Solving for $a$ , you should get:

$a ~= 15.0$

We can now use the fact that the angles in every triangle are supplementary, which means that they have a sum of $180˚$ , to determine the measure of angle B.

$13˚ + 22˚ + B = 180˚$

$B = 180˚ - 22˚ - 13˚$

$B = 145˚$

We can now use this information to reapply The Law of Sines to determine the length of side $b$ .

$sinC/c = sinB/b$

$(sin13˚)/9 = (sin145˚)/b$

Solving for $b$ , we obtain $b ~= 22.9$

In summary:

We have solved the triangle. The measures we have found are as follows.

$a~= 15.0" units"$
$B = 145˚$
$b~= 22.9" units"$

Hopefully this helps!

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How do you solve the triangle given m∠C = 13°, m∠A = 22°, c = 9?

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