In triangle ABC, a=9, c=5, and angle B = 120 degrees. What is the measure of angle A to the nearest degree?

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In triangle ABC, a=9, c=5, and angle B = 120 degrees. What is the measure of angle A to the nearest degree?

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Answer 1 · 2017-11-01T01:13:55

About 39 degree.

Explanation:

First, calculate the edge $b$ $b$ with the law of cosine.

$b^{2} = a^{2} + c^{2} - 2 a c cos B$ $b^2=a^2+c^2-2ac cosB$
$= 9^{2} + 5^{2} - 2 \cdot 5 \cdot 9 \cdot (- \frac{1}{2}) = 151$ $=9^2+5^2-2*5*9*(-1/2)=151$
$b = \sqrt{151}$ $b=sqrt(151)$

Then, calculate the angle $A$ $A$ with the law of sine.
$\frac{a}{sin A} = \frac{b}{sin B}$ $a/(sinA) = b/(sinB)$
$a sin B = b sin A$ $asinB=bsinA$
$sin A = \frac{a sin B}{b} = \frac{9 \cdot \frac{\sqrt{3}}{2}}{\sqrt{151}}$ $sinA=(asinB)/b=(9*sqrt(3)/2)/sqrt(151)$
$= \frac{9 \sqrt{453}}{302}$ $=(9sqrt(453))/302$
$≒0.6343$

Note that $A$ must be smaller than $180-120=60$ degree.
Thus, angle $A$ is
$A=arcsin(0.6343)≒39.37 deg$ .

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In triangle ABC, a=9, c=5, and angle B = 120 degrees. What is the measure of angle A to the nearest degree?

1 Answer

Explanation:

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