Question

in triangle ABC, if tan A/2=5/6,tanC/2=2/5,then. options) a) a, c, b are in A. P. b) a, b, c are in A. P. c) b, a, c are in A. P. d) a, b,c are in G. P​

Answer 1

Answer:

Option (b) a, b, c are in AP

Step-by-step explanation:

if ABC is a triangle then side opposite to angle A is a, side opposite to angle B is b and the side opposite to angle C is c

We know that

$\boxed{\tan\frac{A}{2} =\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}}$

$\boxed{\tan\frac{B}{2} =\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}}$

$\boxed{\tan\frac{C}{2} =\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}}$

where s is the half perimeter

Thus

$\frac{5}{6} =\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}}$

And

$\frac{2}{5} =\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$

$\implies \frac{5}{6}\times\frac{2}{5}  =\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\times \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$

$\implies \frac{1}{3}={\frac{(s-b)}{s}$

$\implies \frac{1}{3}=\frac{(a+b+c)/2-b}{(a+b+c)/2}$

$\implies \frac{1}{3}=\frac{(a-b+c)}{(a+b+c)}$

$\implies a+b+c=3(a-b+c)$

$\implies 2a-4b+2c=0$

$\implies a-2b+c=0$

$\implies a+c=2b$

Therefore, a, b, c are in AP

Answer 2

Answer:

answer is a,b,c in arithmetic progression.q