sin{(b+c)÷2}=cos(a÷2)
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STEP BY STEP,
SUM OF ANGLE OF TRIANGLE = 180
ANGLE A + ANGLE B + ANGLE C = 180
ANGLE B + ANGLE C = 180 - ANGLE A
DIVIDE THIS BY 2
(B+C)/2 = 90 - A/2
TAKING SIN ON BOTH SIDES
SIN(B+C)/2 = SIN( 90 - A/2)
SIN(B+C)/2 = COS A/2 -( SIN 90-A = COS A)
I HOPE THIS WILL HELP YOU...
PLZ MARK AS BRAINLIEST.
SUM OF ANGLE OF TRIANGLE = 180
ANGLE A + ANGLE B + ANGLE C = 180
ANGLE B + ANGLE C = 180 - ANGLE A
DIVIDE THIS BY 2
(B+C)/2 = 90 - A/2
TAKING SIN ON BOTH SIDES
SIN(B+C)/2 = SIN( 90 - A/2)
SIN(B+C)/2 = COS A/2 -( SIN 90-A = COS A)
I HOPE THIS WILL HELP YOU...
PLZ MARK AS BRAINLIEST.
tanishq972003:
hey there is also a another method want to know?
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