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If an angle A of a ΔABC satisfies  5\cos A + 3=0, then the roots of the
quadratic equation, 9x^{2}+27x+20=0
are :

  • Option 1)

    secA, cotA

  • Option 2)

    sinA, secA

  • Option 3)

    secA, tanA

  • Option 4)

    tanA, cosA

 

Answers (2)

best_answer

AS we have learned

Trigonometric Ratios of Functions -

\sin \Theta = \frac{Opp}{Hyp}

\cos \Theta = \frac{Base}{Hyp}

\tan \Theta = \frac{Opp}{Base}

- wherein

Trigonometric Ratios of Functions

 

 

5 \cos A +3= 0\Rightarrow \Cos A = -3/5 

 

on solving 9x^{2}+27x+20=0

x= -4/3 and x= -5/3

so x= \sec A; x = \tan A

 

 

 

 

 

 

 


Option 1)

secA, cotA

This is incorrect 

Option 2)

sinA, secA

This is incorrect 

Option 3)

secA, tanA

This is correct 

Option 4)

tanA, cosA

This is incorrect 

Posted by

Himanshu

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