How do you solve a triangle given <A=84.2°, <B=20.7°, B=17.2?

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How do you solve a triangle given <A=84.2°, <B=20.7°, B=17.2?

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Answer 1 · 2016-04-24T22:18:10

The following diagram represents your problem

Explanation:

enter image source here

Step 1:

Since we know an angle opposite a known side, we can use the Sine Rule to solve for side a.

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

$\frac{sin 84.2}{a} = \frac{sin 20.7}{17.2} = \frac{sin C}{c}$ $sin84.2/a = sin20.7/17.2 = sinC/c$

$a = \frac{17.2 \times sin 84.2}{sin 20.7}$ $a = (17.2 xx sin84.2)/sin20.7$

$a = 48.4$ $a = 48.4$

Step 2:

We could have done this at first, but let's find angle C. We know that the sum of the three angles in a triangle always equals 180. Therefore, we can set up the following equation:

$C + 20.7 + 84.2 = 180$ $C + 20.7 + 84.2 = 180$

$C = 180 - 20.7 - 84.2$ $C = 180 - 20.7 - 84.2$

$C = 75.1˚$

Step 3:

We must now substitute the value of angle C to find side C.

$SinB/b = sinC/c$

$sin20.7/17.2 = sin75.1/c$

$(17.2 xx sin75.1)/sin20.7 = c$

$47.0 = c$

Summary:

Your triangle has the following dimensions:

$/_A = 84.2˚$
$/_B = 20.7˚$
$/_C = 75.1˚$

$a = 48.4$ units
$b = 17.2$ units
$c = 47.0$ units

Hopefully this helps!

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How do you solve a triangle given <A=84.2°, <B=20.7°, B=17.2?

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