How do you solve the triangle given a=21.5, b=16.7, c=10.3?

Oct 31, 2016

By using the Law of Cosines.
ΔABC has angles of 27.8°, 49.2° and 103°
The sum is 180°

Explanation:

Law of Cosines: ${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos C$

${10.3}^{2} = {21.5}^{2} + {16.7}^{2} - 2 \left(21.5 \cdot 16.6\right) \cos C$
$106.09 = 462.25 + 278.89 - 2 \left(356.9\right) \cos C$
$106.09 - 741.14 = - 713.8 \cos C$
$\frac{- 635.05}{-} 713.8 = \cos C$
$C = {\cos}^{-} 1 \left(0.8896749789857103\right)$ ≈ 27.83°

${b}^{2} = {a}^{2} + {c}^{2} - 2 a c \cdot \cos B$
${16.7}^{2} = {21.5}^{2} + {10.3}^{2} - 2 \left(21.5 \cdot 10.3\right) \cdot \cos B$
$278.89 = 462.25 + 106.09 - 2 \left(221.45\right) \cdot \cos B$
$278.89 - 568.34 = - 442.9 \cdot \cos B$
$\frac{- 289.45}{-} 442.9 = \cos B$
$B = {\cos}^{-} 1 \left(0.6535335290133213\right)$≈ 49.19°

${a}^{2} = {b}^{2} + {c}^{2} - 2 b c \cdot \cos C$
${21.5}^{2} = {16.7}^{2} + {10.3}^{2} - 2 \left(16.7 \cdot 10.3\right) \cdot \cos A$
$462.25 = 278.89 + 106.09 - 2 \left(172.01\right) \cdot \cos A$
$462.25 - 384.98 = - 344.02 \cdot \cos A$
$- \frac{77.27}{344.02} = \cos A$
$A = {\cos}^{-} 1 \left(- 0.2246090343584675\right)$≈ 102.98°

So, ΔABC has angles of 27.8°, 49.2° and 103°
The sum is 180°