∆ABC में यदि tanA=1,tanB=2 तोa:b:c=?
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tan(A+B)= tanA+ TanB /1-tanA tanB
= 1+2/1-2
=-3
but A+B+C = π
A+B = π-C
tan (π-c)= -3
tanC = 3
ratio tan~1:tan~2:tan~3
= 1+2/1-2
=-3
but A+B+C = π
A+B = π-C
tan (π-c)= -3
tanC = 3
ratio tan~1:tan~2:tan~3
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