Question

In any triangle ABC, prove that
$\binom{a}{ \sin(capital \: a) } = \binom{b}{ \sin(capital \: b) } = \binom{c}{ \sin(capital \: c) }$
*This question is based on Vectors*​

Answer 1

Answer:

The proof is explained below :

Step-by-step explanation:

Using Sine rule in a triangle ,

Now, LHS = a·sinA - b·sinB

= k·sin²A - k·sin²B { Using Sine rule equations }

= k·[sin(A + B)·sin(A - B)]

= k·sin(π - C)·sin(A - B) { By angle sum property of a triangle sum of all angles is 180 ⇒ A + B + C = π ⇒ A + B = π - C }

= k·sinC·sin(A - B)

= c·sin(A - B) = R.H.S.

So, LHS = RHS

Hence, a·sinA - b·sinB = c·sin(A - B)