If an angle A of a $\Delta$ABC satisfies $5\cos A+3=0$, then the roots of the quadratic equation, $9x^2+27x+20=0$ are.
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$\sin A, \sec A$
- $\sec A, \tan A$
- $\tan A, \cos A$
- $\sec A, \cot A$
Given 5cos A + 3 = 0, cos A = -3/5. In a triangle, this implies A is in the second quadrant. Thus, sec A = -5/3 and tan A = -4/3. The roots of 9x^2 + 27x + 20 = 0 are found using the quadratic formula: x = (-27 +/- sqrt(729 - 720)) / 18 = (-27 +/- 3) / 18. Roots are -24/18 = -4/3 and -30/18 = -5/3. These match tan A and sec A.
The given quadratic equation is 9 x squared plus 27 x plus 20 equals 0. Using the quadratic formula, the discriminant is 27 squared minus 4 times 9 times 20, which is 729 minus 720, yielding 9. The roots evaluate to (-27 plus 3) divided by 18, which is -5/3, and (-27 minus 3) divided by 18, which is -4/3. From the given equation 5 cos A + 3 = 0, we find cos A = -3/5, and using the identity sec squared A = 1 + tan squared A, sec A evaluates to -5/3 and tan A evaluates to -4/3. The roots of the quadratic equation are therefore sec A and tan A.