Question

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

Answer 1

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°