Relation between circumRadius and Sine Rule is given by

R = \frac{a}{S i n A} = \frac{b}{S i n B} = \frac{c}{S i n C}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

R = \frac{a}{2 S i n A} = \frac{b}{2 S i n B} = \frac{c}{2 S i n C}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

R = \frac{a b}{S i n A S i n B} = \frac{b c}{S i n B S i n C} = \frac{c a}{S i n C S i n A}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

R = \frac{a b c}{S i n A S i n B S i n C}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $R = \frac{a}{2 S i n A} = \frac{b}{2 S i n B} = \frac{c}{2 S i n C}$
Relation between CircumRadius and Sine Rule is given by $R = \frac{a}{2 S i n A} = \frac{b}{2 S i n B} = \frac{c}{2 S i n C}$

Suggest Corrections

Similar questions

Prove: $\frac{sin (A - B)}{sin A sin B} + \frac{sin (B - C)}{sin B sin C} + \frac{sin (C - A)}{sin C sin A} = 0$

\frac{sin (A - C) + 2 sin A + sin (A + C)}{sin (B - C) + 2 sin B + sin (B + C)}

A + B + C = π

sin A + sin B - sin C = 2 sin \frac{A}{2} sin \frac{B}{2} cos \frac{C}{2}

\frac{s i n (A + B) + s i n (A - B)}{c o s (A + B) + c o s (A - B)} = t a n A

\frac{s i n (A - B)}{c o s A c o s B} + \frac{s i n (B - C)}{c o s B c o s C} + \frac{s i n (C - A)}{c o s C c o s A} = 0

\frac{s i n (A - B)}{s i n A s i n B} + \frac{s i n (B - C)}{s i n B s i n C} + \frac{s i n (C - A)}{s i n C s i n A} = 0

s i n^{2} B = s i n^{2} A + s i n^{2} (A - B) - 2 s i n A c o s B s i n (A - B)

c o s^{2} + c o s^{2} B - 2 c o s A c o s B c o s (A + B) = s i n^{2} (A + B)

\frac{t a n (A + B)}{c o t (A - B)} = \frac{t a n^{2} A - t a n^{2} B}{1 - t a n^{2} A t a n^{2} B}

Join BYJU'S Learning Program