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IN A TRIANGLE ABC , tan A/2=5/6 and tan c/2=2/5 then a)a.b.c are in A P b)cos A,cos b, cos C are in AP c)sin a,sin b,sin c are in GP d)none of these
tanA/2 = [(s-b)(s-c)/s(s-a)]1/2 ....................1 tanC/2 = [(s-a)(s-b)/s(s-c)]1/2 ..................2 eq1 * eq2 = tanA/2tanB/2 = [s-b/s] .................3 tanA/2 = 5/6 (given) tanC/2 = 2/5 (given) tanA/2tanB/2 = 5/6*2/5 = 1/3 from eq 3 , (s-b)/s = 1/3 3s - 3b = s 2s = 3b 2(a+b+c)/2 = 3b a+c = 2b a,b,c are in AP approve if u like my ans
tanA/2 = [(s-b)(s-c)/s(s-a)]1/2 ....................1
tanC/2 = [(s-a)(s-b)/s(s-c)]1/2 ..................2
eq1 * eq2 = tanA/2tanB/2 = [s-b/s] .................3
tanA/2 = 5/6 (given)
tanC/2 = 2/5 (given)
tanA/2tanB/2 = 5/6*2/5 = 1/3
from eq 3 , (s-b)/s = 1/3
3s - 3b = s
2s = 3b
2(a+b+c)/2 = 3b
a+c = 2b
a,b,c are in AP
approve if u like my ans
dear GAVEESH a,b,c are in AP .........
dear GAVEESH
a,b,c are in AP .........
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