How do you solve the right triangle ABC given A= 20 degrees C= 104 degrees and c=9?

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How do you solve the right triangle ABC given A= 20 degrees C= 104 degrees and c=9?

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Answer 1 · 2015-07-05T21:05:45

$∠ B = 180^{o} - (20^{o} + 102^{o}) = 56^{0}$ $/_B = 180^o -( 20^o + 102^o) = 56^0$
Using the Law of Sines
$XXXX$ $color(white)("XXXX")$ $a = 3.172416$ $a = 3.172416$
$XXXX$ $color(white)("XXXX")$ $b = 7.689757$ $b= 7.689757$

Explanation:

Ignoring the fact that the question declared that the triangle was a right triangle (obviously impossible with one angle = $102^{o}$ $102^o$ ):

The third angle can be calculated using the knowledge that the sum of the interior angles of any triangle is always $180^{o}$ $180^o$ .

Using the Law of Sines in the form:
$XXXX$ $color(white)("XXXX")$ $\frac{c}{sin (C)} = \frac{a}{sin (A)} = \frac{b}{sin (B)}$ $c/sin(C) = a/sin(A) = b/sin(B)$
we can calculate (for example):
$XXXX$ $color(white)("XXXX")$ $a = sin (A) \cdot \frac{c}{sin (C)} = sin (20^{o}) \cdot \frac{9}{sin (102^{o})}$ $a = sin(A)*c/sin(C) = sin(20^o)*9/sin(102^o)$
and through the magic of hand calculators or spreadsheet functions come up with the indicated answers.

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How do you solve the right triangle ABC given A= 20 degrees C= 104 degrees and c=9?

1 Answer

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