Question

Prove that in any triangle abc sin/a=sinb/b=sinc/c

Answer 1

Answer:

Step-by-step explanation:

We know from property of circles if a chord subtends an angle 2Ф at its center, then it subtends an angle Ф at any other point on the same segment of the circle and vice-versa.

Let AB = c, BC =a and CA =b.

Now ∠BOC will be 2∠A,

∠COA will be 2∠B and

∠AOB will be 2∠C.

If we drop a perpendicular from the center(O) to the chordBC) say the foot be D , we get 2 congruent triangles BOD and COD(R-H-S congruency) , hence ∠BOD = ∠COD =A..

If we assume the radius of the circle to be R,

Now sin∠BOD =( a/2)/R

=>sinA = a/2R

or 2R = a/sinA----(*)

Similarly from triangle COA, we get

2R = b/sinB---(**)

for triangle AOB, we get

2R = c/sinC...(***),

Hence, from (*),(**)and (***), we can conclude that 2R = a/sinA = b/sinB=c/sinC.