Question~ [tex] \bold {If A, B \: and \: C \: are \: interior \: angles \: of \: a \: triangle \: ABC \: then \: show \: that \: \sin (\frac{B +C }{2} ) = \cos \: \frac{A}{2} }[/tex]Don’t spam!! About the author Skylar
♡Answer♡:– Given △ABC We know that sum of three angles of a triangle is 180 Hence ∠A+∠B+∠C=180° or A+B+C=180° B+C=180° −A Multiply both sides by 1/2 1/2 (B+C)= 1/2 (180° −A) 1)2 (B+C)=90° − A/2 …(1) Now 1/2 (B+C) Taking sine of this angle sin( B+C/2) [B+C/2 =90° − A/2 ] sin(90° − A/2 ) cos A/2 [sin(90°−θ)=cosθ] Hence sin( B+C/2 )=cos A/2 proved Itzgirl45 ❤️ Reply
Step-by-step explanation:
Since A+B+C=180 for interior angles of triangle ABC. then B+C=180-A.
♡Answer♡:–
Given △ABC
We know that sum of three angles of a triangle is 180
Hence ∠A+∠B+∠C=180°
or A+B+C=180°
B+C=180° −A
Multiply both sides by 1/2
1/2 (B+C)= 1/2 (180° −A)
1)2 (B+C)=90° − A/2 …(1)
Now 1/2 (B+C)
Taking sine of this angle
sin( B+C/2) [B+C/2 =90° − A/2 ]
sin(90° − A/2 )
cos A/2 [sin(90°−θ)=cosθ]
Hence sin( B+C/2 )=cos A/2
proved
Itzgirl45 ❤️