Question 5 i & ii

  • 1
in a triangle
A+B+C = 180
(A+B+C)/2 = 90
thus
(A+B)/2 = 90 - C/2
(B+C)/2 = 90 - A/2

thus sin((A+B)/2 ) = sin ( 90 -C/2) = cos(C/2)
tan( (B+C)/2 ) = tan ( 90 - A/2 ) = cot(A/2)
  • 0
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