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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