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

1 Answer
J'Neal · EZ as pi
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^2-2abcosC

10.3^2=21.5^2+16.7^2-2(21.5*16.6)cosC
106.09=462.25+278.89-2(356.9)cosC
106.09-741.14=-713.8cosC
(-635.05)/-713.8=cosC
C= cos^-1(0.8896749789857103) ≈ 27.83°

b^2=a^2+c^2-2ac*cosB
16.7^2=21.5^2+10.3^2-2(21.5*10.3)*cosB
278.89=462.25+106.09-2(221.45)*cosB
278.89-568.34=-442.9*cosB
(-289.45)/-442.9=cosB
B=cos^-1(0.6535335290133213)≈ 49.19°

a^2=b^2+c^2-2bc*cosC
21.5^2=16.7^2+10.3^2-2(16.7*10.3)*cosA
462.25=278.89+106.09-2(172.01)*cosA
462.25-384.98=-344.02*cosA
-77.27/344.02=cosA
A=cos^-1(-0.2246090343584675)≈ 102.98°

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

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