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Fernanda Nicholson
Mathematics
25 September, 20:57
Solve triangle ABC, when A = 6, B=10 and c=12
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Jamarion
25 September, 21:04
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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