Question

P is the circumcentre of an acute angle triangle ABC with circumradius R . Midpoint of BC is D .Show that the perimeter of triangle ABC is ( Sin A + sin b + sin c)​

Answer 1

Answer:

Heyaa I think your question is incomplete.

We have to prove :-

The perimeter of triangle ABC is

2R(Sin A + Sin B + Sin C)

And here is the solution :-

Sine Formula :-

In ΔABC:-

$\frac{a}{sina} = \frac{b}{sinb} = \frac{c}{sinc} = 2r$

=> a = 2R sinA

b = 2R sinB

c = 2R sinC

Now, perimeter of ΔABC:-

=> a + b + c

=> 2R sinA + 2R sinB + 2R sinC

=> 2R(SinA + SinB + SinC)

Hence proved.

Hope it helped you.