Question

In \( \triangle ABC \) with usual notations, prove that: \( \frac{a}{\sin A }=\frac{b}{\sin B }=\frac{c}{\sin C }=2 R \), where \( R \) is the circumradius of \( \triangle ABC \).

Answer 1

a = side length opposite to ∠A = BC

b = side length opposite to ∠B = AC

and c = side length opposite to ∠c = AB

O is circumcentre of △ABC.

Construction: (i) Join OB

(ii) Produced OB intersect circle at point D

(iii) Join CD.

\(\because\) BD is diameter of circumcircle of ΔABC whose circurmradius is R.

\(\therefore\) BD  = 2R

∠BCD = 90° (\(\because\) Angle made in semi-circle is 90°)

Also, ∠BAC & ∠BDC are angles in same segment

\(\therefore\) ∠BAC = ∠BDC-----(i)

In right triangle BCD, we have

sin ∠BDC = \(\frac{BC}{BC}=\frac{a}{2R}\)

\(\therefore\) sin ∠BAC = \(\frac{a}{2R}\) (From (i))

⇒ sin A = \(\frac{a}{2R}\)

⇒ \(\frac{a}{sinA}\) = 2R

Similarly, \(\frac{b}{sinB}\) = 2R and \(\frac{c}{sinC}\) = 2R.

Hence, \(\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}\) = 2R, where R is circumradius  of ΔABC.

Hence Proved