Question

P is a circumcenter of acute angle triangle ABC with circumradius R.midpoint of BC is D show that perimeter of triangle ABC is 2R( Sin A + sin b + sin c)

Answer 1

$\textbf{Concept used:}$

$\textsf{Sine formula:}$

$\textsf{In triangle ABC,}$

$\mathsf{\displaystyle\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}=2R}$

$\textsf{where R- circumradius}$

$\text{From sine fromula, we have}$

$a=2R\;sinA$

$b=2R\;sinB$

$c=2R\;sinC$

$\text{Now, Perimeter of \triangle ABC}$

$=\text{sum of all lengths of sides of \triangle ABC}$

$=a+b+c$

$=2R\;sinA+2R\;sinB+2R\;sinC$

$=2R(sinA+sinB+sinC)$

$\therefore\textbf{The perimeter of \triangle ABC is 2R(sinA+sinB+sinC)}$

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