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In a triangle ABC, if cotA :cotB :cotC =...

In a triangle `ABC, if cotA :cotB :cotC = 30: 19 : 6` then the sides `a, b, c` are

A

in A.P.

B

in G.P.

C

in H.P.

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
A

`cot A = (cos A)/(sin A) = ((b^(2) + c^(2) -a^(2))/(2bc))/((a)/(2R)) = ((b^(2) + c^(2) -a^(2))/(abc)) R`
`:. cot A : cot B : cot C`
`= b^(2) + c^(2) -a^(2) : c^(2) + a^(2) - b^(2) : a^(2) + b^(2) -c^(2)`
Clearly if `cot A : cot B : cot C = l : m : n`
then `a: b : c = sqrt(m + n) : sqrt(n +l) : sqrt(l + m)`
`= sqrt(19 + 6) : sqrt(6 + 30) : sqrt(30 + 19)`
`= 5 : 6 : 7`
Therefore, a, b, c are in A.P.
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