In a ΔABC, if 3angleA = 4angleB = 6angleC, find the angles
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Answered by
4
A + B + C = 180
3A = 4B = 6C = K (let)
A = K/3
B = K/4
C = K/6
So, A + B + C = (K/3) + (K/4) + (K/6) = 9K/12 = 3K/4
3K/4 = 180
K = 240
So, A = 80, B = 60, C = 40
3A = 4B = 6C = K (let)
A = K/3
B = K/4
C = K/6
So, A + B + C = (K/3) + (K/4) + (K/6) = 9K/12 = 3K/4
3K/4 = 180
K = 240
So, A = 80, B = 60, C = 40
manasi18:
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Answered by
2
A+B+C=180
GIVEN 3A=4B=6C
LET IT BE EQUAL TO S
A=S/3,B=S/4,C=S/6
S/3+S/4+S/6=180
4S+3S+2S/12=180
S=240
A=80,B=60,C=40
GIVEN 3A=4B=6C
LET IT BE EQUAL TO S
A=S/3,B=S/4,C=S/6
S/3+S/4+S/6=180
4S+3S+2S/12=180
S=240
A=80,B=60,C=40
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