Question

in a triangle abc if 3a=bc then what is the value of CotA/2 CotB/2

Answer 1

Let a<=b<=c. If not, relable them so that this is so. Since a+D=b and b+D=c, it follows that 2b=a+c. We use the law of cotangents. See http:// en.wikipedia .org/wiki/ Law_of_ cotangents

Then cot(A/2) = (s-a)/r, cot(B/2) = (s-b)/r and cot(C/2) = (s-c)/r where s denotes the semiperimeter (a+b+c)/2 and r denotes the radius of the circle that can be inscribed in triangle ABC.

It follows that cot(A/2) / s-a = cot(B/2) / s-b = cot(C/2) / s-c.

As before, if cot(A/2), cot(B/2), and cot(C/2) are in AP it follows that cot(B/2) = [cot(A/2) + cot(C/2)]/2. But this follows from the previous formulas since

[cot(A/2) + cot(C/2)]/2 = [((s-a)/(s-b)) cot(B/2) + ((s-c)/(s-b)) cot(B/2)]/2

=(1/2) * ((2s - a -c )/(s-b) ) cot (B/2)

=(1/2) * ((a+ 2b +c)/(a+c)) cot(B/2) using 2s=a+b+c

=(1/2) * ((2a+2c)/(a+c)) cot (B/2) using that 2b=a+c by hypothesis

=cot (B/2) as desired.

Please mark as brainlist answer

Please mark as brainlist answer

Answer 2

Answer:

2 is the answer.

Step-by-step explanation:

given,3a=b+c

2s=a+b+c

=>s=a+b+c/2=a+3a/2=4a/2 [putting b+c=3a]

=>s=2a

cotB×cotC=1/tanB×1/tanC

=whole root over of s(s-b)/(s-a)(s-b)×whole root over of s(s-c)×(s-a)(s-b)

=whole root over of s^2(s-b)(s-c)/(s-a)^2(s-b)(s-c)

=whole root over of s^2/(s-a)^2

=s/s-a

=2a/2a-a

=2a/a

=2(answer)