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Question

If an angle A of a ABC satisfies 5cosA+3=0, then the roots of the quadratic equation, 9x2+27x+20=0 are:

A
secA,cotA
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B
secA,tanA
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C
tanA,cosA
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D
sinA,secA
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Solution

The correct option is B secA,tanA
5cosA+3=0cosA=35 this means A(90,180)
So,secA=1cosA=53sinA=1cos2A=45tanA=sinAcosA=43
Now roots of the equation 9x2+27x+20=09x2+12x+15x+20=03x(3x+4)+5x(3x+4)=0(3x+5)(3x+4)=0
So, x= 53 and 43
So roots are secA and tanA.

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