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Question

If 2cosA+3cosB+5cosC=2sinA+3sinB+5sinC=0 then
8cos3A+27cos3B+125cosC=kcos(A+B+C) then k=

A
70
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B
80
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C
90
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D
60
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Solution

The correct option is C 90
2cosA+3cosB+5cosC=2sinA+3sinB+5sinC=0

cosθ=12(ejθ+ejθ)

sinθ=12j(ejθejθ)

212(ejA+ejA)+312(ejB+ejB)+512(ejC+ejC)=0
2(ejA+ejA)+3(ejB+ejB)+5(ejC+ejC)=0(i)

Also,2(ejAejA)+3(ejBejB)+5(ejCejC)=0(ii)

As(i)(ii)

4ejA+6ejB+10ejC=0

2ejA+3ejB+5ejC=0

Cubing on both sides.
8ej3A+27ej3B+125ej3C+3(ejA×2×3ejB.5ejC)=0

8ej3A+27ej3B+125ej3C=90ejAejBejC
k=90

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