if, sinA+sinB+sinC=3 then find cosA+cosB+cosC
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Answered by
58
Maximum value of sin Ф is 1.
So sinA+sinB+sinC=3 means
sin A = 1
sin B = 1
sin C = 1
Thus, A = B = C = 90 degree
cos A = cos 90 = 0
cos B = cos 90 = 0
cos C = cos 90 = 0
cos A + cos B +cos C = 0
So sinA+sinB+sinC=3 means
sin A = 1
sin B = 1
sin C = 1
Thus, A = B = C = 90 degree
cos A = cos 90 = 0
cos B = cos 90 = 0
cos C = cos 90 = 0
cos A + cos B +cos C = 0
Answered by
8
Answer:
ans is : 0. zero
Step-by-step explanation:
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