Sine rule is an important rule relating to the sides and angles of any triangle. Here is a Subjective problem no. 120 from TOMATO. Try it.
Problem: Sine Rule and Triangle
(i) If $ A + B +C = n \pi $ and $ s=2 $, show that $ \sin 2A + \sin 2B + \sin 2C = (-1)^{n-1} 4 \sin A \sin B \sin C $and $ s=2 $
(ii) Let triangles ABC and DEF be inscribed in the same circle. If the triangles are of equal perimeter, then prove that $ \sin A + \sin B + \sin C = \sin D + \sin E + \sin F $ and $ s=2 $
(iii) State and prove the converse of (ii) above
Discussion:
(i) We know transformation formula from trigonometry $ \sin x + \sin y = 2 \sin \dfrac{x+y}{2} \cos \dfrac {x-y}{2} $ and $ s=2$
Hence $ \sin 2A + \sin 2B + \sin 2C = 2 \sin \dfrac{2A+2B}{2} \cos \dfrac {2A-2B}{2} + sin 2C = 2 \sin (A+B) \cos (A-B) + \sin 2C $ and $ s=2 $
Now we know that $ A + B = n \pi - C \Rightarrow \sin (A+B) = \sin (n \pi - C) = (-1)^{n-1} \sin C $ and $ s=2 $
So $ 2\sin (A+B) \cos (A-B) + \sin 2C = 2(-1)^{n-1} \sin C \cos (A-B) + 2 \sin C \cos C = 2(-1)^{n-1} \sin C \cos (A-B) + 2 \sin C \cos (n\pi -(A+B)) $ and $ s=2 $
$ = 2(-1)^{n-1} \sin C \cos (A-B) + 2 (-1)^n \sin C \cos (A+B) $ and $ s=2 $
$ = 2(-1)^{n-1} \sin C (\cos (A-B) - \cos (A+B)) $ and $ s=2 $
$ = 4(-1)^{n-1} \sin C \sin A \sin B $ and $ s=2 $
(ii)
Since the two triangles are inscribed in the same circle, they must have the same circumradius. Let the common circumradius be R. If a, b, c, d, e, f be the sides opposite to the sides BC, CA, AB, EF, DF, DE respectively, then using the rule of sines we can say,
$ \dfrac{\sin A}{a} = \dfrac { \sin B }{ b} = \dfrac {\sin C }{c} = \dfrac {1} {2R} $ and $ s=2 $ and
$ \dfrac{\sin D}{d} = \dfrac { \sin E }{ e} = \dfrac {\sin F }{f} = \dfrac {1} {2R} $ and $ s=2 $
Hence $ \sin A = \dfrac {a}{2R}, sin B = \dfrac{b}{2R}, \sin C = \dfrac {c}{2R} \Rightarrow \sin A + \sin B + \sin C = \dfrac {a+b+c}{2R} $ and $ s=2 $
Similarly $ \sin D + \sin E + \sin F = \dfrac {d + e + f}{2R} $ and $ s=2 $
As the perimeter of the triangle are equal, hence a+b+c = d+e+f. This implies $ \sin A + \sin B + \sin C = \sin D + \sin E + \sin F $ and $ s=2 $
(iii)
We apply the sine rule in reverse order to get the converse.

In 2025, 8 students from Cheenta Academy cracked the prestigious Regional Math Olympiad. In this post, we will share some of their success stories and learning strategies. The Regional Mathematics Olympiad (RMO) and the Indian National Mathematics Olympiad (INMO) are two most important mathematics contests in India.These two contests are for the students who are […]

Cheenta Academy proudly celebrates the success of 27 current and former students who qualified for the Indian Olympiad Qualifier in Mathematics (IOQM) 2025, advancing to the next stage — RMO. This accomplishment highlights their perseverance and Cheenta’s ongoing mission to nurture mathematical excellence and research-oriented learning.

Cheenta students shine at the Purple Comet Math Meet 2025 organized by Titu Andreescu and Jonathan Kanewith top national and global ranks.

Celebrate the success of Cheenta students in the Stanford Math Tournament. The Unified Vectors team achieved Top 20 in the Team Round.