if a b and c are interior angles of a triangle abc then find tan(b+c/2)
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hiHi☺☺
a, b, c are interior angles of triangle,
•°• a+b+c = 180°
b+c = 180°-a
Dividing both side by 2
we get,
(b+c)/2 = 90°-(a/2)
=> tan{(b+c)/2} = tan (90-a/2)
°•° tan(90-x) = cotx
•°• tan{(b+c)/2} = cot(a/2) .
☺☺
a, b, c are interior angles of triangle,
•°• a+b+c = 180°
b+c = 180°-a
Dividing both side by 2
we get,
(b+c)/2 = 90°-(a/2)
=> tan{(b+c)/2} = tan (90-a/2)
°•° tan(90-x) = cotx
•°• tan{(b+c)/2} = cot(a/2) .
☺☺
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