show that in Δ ABC: sin A+B/2=cos C/2
Answers
Answered by
1
Hey!Here is your answer..
Angle A +Angle B +Angle C =180°
So,L.H.S-> sin A+B/2
= sin{(180°-C)/2}
=sin (90°-C/2)
=cos C/2=>R.H.S
Angle A +Angle B +Angle C =180°
So,L.H.S-> sin A+B/2
= sin{(180°-C)/2}
=sin (90°-C/2)
=cos C/2=>R.H.S
isharao:
thank you so much, really kind of you!
Answered by
1
A+B+C=180degree(angle sum property of triangle)
B+A=180°-C
dividing above equation by 2
B+A/2=180/2-C/2
B+A/2=90°-C/2
multiply above equation by sin
sin(B+A/2)=sin(90°-C/2)
therefore
sin A+B/2=cos C/2
B+A=180°-C
dividing above equation by 2
B+A/2=180/2-C/2
B+A/2=90°-C/2
multiply above equation by sin
sin(B+A/2)=sin(90°-C/2)
therefore
sin A+B/2=cos C/2
thank you so much.
pls select my answer as brainliest
Similar questions