If A,B,C are the interior angle of a triangle ABC, show that :
cos A+B/2 = Sin C/2
Please someone help me to do this sum
Answers
Answered by
0
Answer:
Given △ABC
We know that sum of three angles of a triangle is 180
Hence ∠A+∠B+∠C=180
o
or A+B+C=180
o
B+C=180
o
−A
Multiply both sides by
2
1
2
1
(B+C)=
2
1
(180
o
−A)
2
1
(B+C)=90
o
−
2
A
...(1)
Now
2
1
(B+C)
Taking sine of this angle
sin(
2
B+C
)[
2
B+C
=90
o
−
2
A
]
sin(90
o
−
2
A
)
cos
2
A
[sin(90
o
−θ)=cosθ]
Hence sin(
2
B+C
)=cos
2
A
proved
Answered by
0
Answer:
If A, B, C are the interior angles of triangle ABC, then show that cos(B+C/2)= sin A/2 As we know, for any given triangle, the sum of all its interior angles is equals to 180°. Now, put sin function on both sides.
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