Question

In ΔABC, show that ∑a (sin B - sin C) = 0.

Answer 1

Answer:

Answer is 0

Step-by-step explanation:

I have applied sine formula to solve this problem.

In any triangle ABC

(a/ sinA ) = (b/ sinB) = (c/ sinC) = 2R

where R is the circum radius

a( sinB - sinC) + b (sinC - sinA)

+ c(sinA - sinB)

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

= 2R sinAsinB - 2R sinA sinC

+ 2R sinB sinC - 2R sinB sinA

+ 2R sinC sinA - 2R sinC sinB

=2R sinAsinB - 2R sinC sinA

+ 2R sinB sinC - 2R sinA sinB

+ 2R sinC sinA - 2R sinB sinC

=0

I hope this answer helps you