If A,B and C are the interior
angles of a AABC, then what
is the value of Cos(A+B)/2
Answers
Answered by
0
Given, A, B, C are the interior angles of a triangle ABC
∴∠A+∠B+∠C=180
∴
2
∠A+∠B+∠C
=
2
180
∴
2
∠A+∠B+∠C
=90
∴
2
∠A+∠B
=90−
2
∠C
Now, taking cos on both sides. we get,
cos(
2
∠A+∠B
)=cos(90−
2
∠C
)
⇒cos(
2
∠A+∠B
)=sin(
2
∠
∴∠A+∠B+∠C=180
∴
2
∠A+∠B+∠C
=
2
180
∴
2
∠A+∠B+∠C
=90
∴
2
∠A+∠B
=90−
2
∠C
Now, taking cos on both sides. we get,
cos(
2
∠A+∠B
)=cos(90−
2
∠C
)
⇒cos(
2
∠A+∠B
)=sin(
2
∠
Answered by
1
Answer:
Sorry only for point please sorry
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