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  2. Can someone explain - Trigonometry - JEE Main
# Engineering

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)
A Anuj Pandey
H Himanshu

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 

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