How do you solve the triangle ABC given a=55, b=25, c=72?

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Answer 1 · 2016-08-24T17:09:34

$hatA = 39.3°, hatB = 16.8° , hatC = 123.9°$

"Solve a triangle" means find all the missing information - in this case, the size of each angle.

The lengths of 3 sides are given. We must use the Cosine Rule.

Find the biggest angle first - if it is an obtuse angle, the Cos rule will indicate this, but the Sin Rule will not,

Angle C is the biggest angle because it is opposite the longest side.

$Cos hatC = (a^2 + b^2 - c^2)/(2ab) = (55^2 +25^2 -72^2)/(2xx55xx25)$

$Cos hatC = -0.5578" (the negative value means it is obtuse)"$

$color(red)(hatC =123.9°)$

Now use the SIn Rule.

$(Sin hatA)/a = (Sin hatC)/c$

$Sin hatA = (55 xx Sin 123.9)/72 = 0.6340$

$color(red)(hatA = 39.3°)$

We have 2 angles, use the angle sum of a triangle to find $hatB$

$hatB = 180° - 123.9° - 39.3°$

$color(red)(hatB = 16.8°)$