Solve triangle ABC, when A= 6, B=10 and c=12

1 answer:

5 0

Use cosine rule,

cos(A)=(b^2+c^2-a^2)/(2bc)
=(10^2+12^2-6^2)/(2*10*12)
=13/15
A=29.926 degrees.................................(A)

cos(B)=(c^2+a^2-b^2)/(2ca)
=(12^2+6^2-10^2)/(2*12*6)
=5/9
B=56.251 degrees.................................(B)

cos(C)=(a^2+b^2-c^2)/(2ab)
=(6^2+10^2-12^2)/(2*6*10)
=-1/15
C=93.823 degrees.................................(C)

Check:29.926+56.251+93.823=180.0 degrees....ok

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