If an angle A of a `triangle ABC` satisfies `5cos A + 3 = 0,` then the roots of the quadratic equation, `9x^2 + 27x + 20 = 0` are

Options

A. sec A, secA
B. sec A, cot A
C. sec A, tan A
D. tan A, cos A

Answer

To solve the problem step by step, we need to follow these steps:

### Step 1: Solve for cos A
Given the equation:
\[ 5 \cos A + 3 = 0 \]

We can isolate cos A:
\[ 5 \cos A = -3 \]
\[ \cos A = -\frac{3}{5} \]

### Step 2: Find sec A
The secant function is the reciprocal of the cosine function:
\[ \sec A = \frac{1}{\cos A} \]
Substituting the value we found:
\[ \sec A = \frac{1}{-\frac{3}{5}} = -\frac{5}{3} \]

### Step 3: Use the quadratic formula
We need to find the roots of the quadratic equation:
\[ 9x^2 + 27x + 20 = 0 \]

We will use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 9 \), \( b = 27 \), and \( c = 20 \).

### Step 4: Calculate the discriminant
First, we calculate the discriminant \( b^2 - 4ac \):
\[ b^2 = 27^2 = 729 \]
\[ 4ac = 4 \times 9 \times 20 = 720 \]
Thus,
\[ b^2 - 4ac = 729 - 720 = 9 \]

### Step 5: Substitute values into the quadratic formula
Now we can substitute the values into the quadratic formula:
\[ x = \frac{-27 \pm \sqrt{9}}{2 \times 9} \]
\[ x = \frac{-27 \pm 3}{18} \]

### Step 6: Calculate the two roots
Calculating the two roots:
1. For the positive root:
\[ x_1 = \frac{-27 + 3}{18} = \frac{-24}{18} = -\frac{4}{3} \]

2. For the negative root:
\[ x_2 = \frac{-27 - 3}{18} = \frac{-30}{18} = -\frac{5}{3} \]

### Final Answer
The roots of the quadratic equation \( 9x^2 + 27x + 20 = 0 \) are:
\[ x = -\frac{4}{3}, -\frac{5}{3} \]