Question

If B+C=60' prove that sin(120'-B) =sin(120'-C)

Answer 1

Hello!

L.H.S. = sin ( 120° - B )

= sin ( 120° - ( 60° - C ) )

= sin ( 120° - 60° + C )

= sin ( 60 + C )

= sin ( 180° - ( 60° + C ) )

= sin ( 120° - C ) = R.H.S

$\boxed{\sf HENCE\:PROVED}$

Cheers!

Answer 2

Answer:

It is proven below

Step-by-step explanation:

Given that B+C =60°

                 B =60°- C  

here we need to prove that sin(120°- B) = sin(120°-C)

take LHS ,  

⇒ sin (120°- B )

⇒ sin (120° -( 60° -C)      (from above data)

⇒ sin (120° - 60° + C)  

⇒ sin ( 60° + C)    

⇒ sin(180°-(60°+C))    [ sin (180° – θ) = sin θ for any value of  θ ]

⇒ sin(180° - 60°-C)    

⇒ sin(120°-C) = RHS

hence it is proven that  if B+C= 60° then

                          sin(120°-B) = sin(120°-C)