If an angle A of a ΔABC satisfies 5 cos A + 3 = 0, then the roots of the quadratic equation, 9x² + 2

Q: Wait, why is angle A necessarily in the second quadrant?

In any triangle $ABC$, the sum of angles is $\pi$ radians. If $\cos A$ were positive, $A$ would be acute. But since $\cos A = -\frac{3}{5}$, the angle must be obtuse. An obtuse angle in a triangle must lie between $\frac{\pi}{2}$ and $\pi$, which is the definition of the second quadrant.

Q: Can we use $\sec^2 A = 1 + \tan^2 A$ instead of drawing a triangle?

Absolutely! You can use identities to find the values, but drawing the triangle is a powerful tool to avoid sign errors. It keeps your geometric intuition sharp and ensures you don't lose track of the negative signs in the second quadrant.

Q: Why do the roots of the quadratic equation turn out to be negative?

The quadratic equation $9x^2 + 27x + 20 = 0$ has coefficients that are all positive. By the properties of quadratic equations, if all coefficients are positive, the roots must be negative. This aligns perfectly with our trigonometric values, as both $\sec A$ and $\tan A$ are negative in the second quadrant.

Question

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

Options:

(A) $\sin A, \sec A$

(B) $\sec A, \cot A$

(C) $\sec A, 	an A$

(D) $	an A, \cos A$

Sigma Percentile · Accepted Answer

### The Harmony of Algebra and Geometry

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to explore a problem that might seem like two separate puzzles at first glance: a trigonometric condition and a quadratic equation.

But as we peel back the layers, you will see how they dance together in perfect harmony. Let us embark on this journey.

### Phase 1: The Trigonometric Detective

We begin with the condition $5 \cos A + 3 = 0$. Our first instinct, as always, is to isolate the unknown.

By transposing the $3$ and dividing by $5$, we find that:

$$\cos A = -\frac{3}{5}$$

Stop right there! Do not just write this down and move on. Look at that negative sign.

In the world of trigonometry, a negative cosine value is a massive clue. Since $A$ is an angle of a triangle, it must be between $0$ and $\pi$.

If $\cos A$ were positive, $A$ would be acute. But here, it is negative, which forces $A$ into the second quadrant, where $A \in (\frac{\pi}{2}, \pi)$. This is the first piece of our puzzle.

### Phase 2: The Geometry of the Second Quadrant

Now, let us visualize this. Imagine a right-angled triangle in the second quadrant.

We know that $\cos A = \frac{	ext{Base}}{	ext{Hypotenuse}} = -\frac{3}{5}$. We can set our base to $-3$ and our hypotenuse to $5$.

Using the timeless Pythagoras theorem, we calculate the height:

$$	ext{Height} = \sqrt{5^2 - (-3)^2} = \sqrt{25 - 9} = \sqrt{16} = 4$$

With this, we can easily find the other ratios. $\sec A$ is the reciprocal of $\cos A$, giving us $-\frac{5}{3}$.

And $	an A$? It is the ratio of height to base, which is:

$$	an A = \frac{4}{-3} = -\frac{4}{3}$$

We have successfully decoded the trigonometric side of our problem.

### Phase 3: The Algebraic Challenge

Now, we shift gears to the quadratic equation: $9x^2 + 27x + 20 = 0$. We need to find its roots.

To factorize this, we look for two numbers that add up to $27$ and multiply to $9 	imes 20 = 180$. After a moment of thought, we find the numbers $15$ and $12$.

Splitting the middle term, we rewrite the equation as:

$$9x^2 + 15x + 12x + 20 = 0$$

Grouping the terms, we get $3x(3x + 5) + 4(3x + 5) = 0$, which simplifies to $(3x + 5)(3x + 4) = 0$.

Setting each factor to zero, we find the roots:

$$x = -\frac{5}{3} \quad 	ext{and} \quad x = -\frac{4}{3}$$

### Phase 4: The Grand Synthesis

Look at what we have achieved! Our trigonometric analysis gave us $\sec A = -\frac{5}{3}$ and $	an A = -\frac{4}{3}$.

Our algebraic analysis gave us the roots $-\frac{5}{3}$ and $-\frac{4}{3}$. They are identical!

The roots of the quadratic equation are exactly **$\sec A$** and **$	an A$**.

This is the beauty of mathematics—when two seemingly unrelated concepts converge to reveal a single, elegant truth. You have not just solved a problem; you have witnessed the interconnectedness of the mathematical universe. Keep this curiosity alive, and you will conquer any challenge the JEE throws your way!

Animated Solution for Mathematics - Trigonometry: If an angle A of a $△ A B C$ satisfies $5 cos A + 3 = 0$ , then the roots of the quadratic equation, $9 x^{2} + 27 x + 20 = 0$ are :

Select Answer:

Visualized Solution

Analyze the Given Condition

Determine the Quadrant of $A$

Geometric Representation

Calculate $sec A$ and $tan A$

Analyze the Quadratic Equation

Factorize the Equation

Solve for the Roots

Final Conclusion

The Sigma Insight: Trigonometric Ratios and Identities

The Harmony of Algebra and Geometry

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If an angle A of a △ABC satisfies 5cosA+3=0, then the roots of the quadratic equation, 9x2+27x+20=0 are :