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 BsecA,tanA 5cosA+3=0⇒cosA=−35 this means A∈(90∘,180∘) So,secA=1cosA=−53sinA=√1−cos2A=45⇒tanA=sinAcosA=−43 Now roots of the equation 9x2+27x+20=0⇒9x2+12x+15x+20=03x(3x+4)+5x(3x+4)=0(3x+5)(3x+4)=0 So, x= −53and−43 So roots are secAandtanA.