How do you solve a triangle if you are given a = 11, b = 14, c = 20?

Jun 5, 2015

Have a look (and check my math):

So:
$h = 7.41 = 14 \sin \left(\alpha\right)$ then alpha=31.95°
$h = 7.41 = 11 \sin \left(\beta\right)$ then beta=42.35°
gamma=180-31.95-42.35=105.70°

Jun 8, 2015

Use the trig identity: c^2 = a^2 + b^2 - 2ab.cosC

400 = 121 + 196 - 308.cos A

$\cos A = - \frac{83}{308} = - 0.27 \to A = 105.66$

Next, find cos B.

196 = 400 + 121 - 440.cos B

$\cos B = \frac{325}{440} = 0.74 - \to B = 42.38$ deg

C = 180 - A - B = 180 - 105.66 - 42.38 = 31.96
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