Complex Numbers And Quadratic Equations question 183
Question: In a triangle $ ABC $ the value of $ \angle A $ is given by $ 5\cos A+3=0 $ , then the equation whose roots are $ \sin A $ and $ \tan A $ will be [Roorkee 1972]
Options:
A) $ 15x^{2}-8x+16=0 $
B) $ 15x^{2}+8x-16=0 $
C) $ 15x^{2}-8\sqrt{2}x+16=0 $
D) $ 15x^{2}-8x-16=0 $
Show Answer
Answer:
Correct Answer: B
Solution:
Given that $ 5\cos A+3=0 $ or $ \cos A=-\frac{3}{5} $ Let $ \alpha =\sin A $ and $ \beta =\tan A $ , then the sum of roots $ =\alpha +\beta =\sin A+\tan A $ $ =\sin A+\frac{\sin A}{\cos A}=\frac{\sin A}{\cos A}(1+\cos A) $ $ =\frac{\sqrt{1-9/25}}{-3/5}( 1-\frac{3}{5} )=\frac{4}{-5}.\frac{5}{3}.\frac{2}{5}=\frac{8}{-15} $ and product of roots $ \alpha .\beta =\sin A\tan A=\frac{{{\sin }^{2}}A}{\cos A} $ $ =\frac{16/25}{-3/5}=-\frac{16}{25}\times \frac{5}{3}=-\frac{16}{15} $ Thus required equation is $ x^{2}-(\alpha +\beta )x+\alpha \beta =0 $
Þ $ x^{2}+\frac{8x}{15}-\frac{16}{15}=0 $
Þ $ 15x^{2}+8x-16=0 $
BETA
