In a triangle `ABC, /_A=x^(2),/_B=y^(@)` and `/_C=(y+20)^(@)`. If `4x+y=10`, then the triangle is

Options

A. Right angled
B. Obtuse angled
C. Equilateral
D. None of these

Answer

To solve the problem, we need to find the nature of triangle ABC given the angles and the equation involving x and y.

### Step 1: Set up the equations based on the angles of the triangle.
The angles of triangle ABC are given as:
- Angle A = \( x^2 \)
- Angle B = \( y \)
- Angle C = \( y + 20 \)

According to the triangle angle sum property, the sum of the angles in a triangle is \( 180^\circ \). Therefore, we can write:
\[ x^2 + y + (y + 20) = 180 \]
This simplifies to:
\[ x^2 + 2y + 20 = 180 \]
Subtracting 20 from both sides gives us:
\[ x^2 + 2y = 160 \quad \text{(Equation 1)} \]

### Step 2: Use the second equation provided.
We are also given the equation:
\[ 4x + y = 10 \quad \text{(Equation 2)} \]

### Step 3: Solve for y in terms of x using Equation 2.
From Equation 2, we can express \( y \) in terms of \( x \):
\[ y = 10 - 4x \]

### Step 4: Substitute y in Equation 1.
Now, substitute \( y \) from Equation 2 into Equation 1:
\[ x^2 + 2(10 - 4x) = 160 \]
This simplifies to:
\[ x^2 + 20 - 8x = 160 \]
Rearranging gives:
\[ x^2 - 8x + 20 - 160 = 0 \]
\[ x^2 - 8x - 140 = 0 \quad \text{(Equation 3)} \]

### Step 5: Solve the quadratic equation using the quadratic formula.
We can solve Equation 3 using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -8, c = -140 \):
\[ x = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot (-140)}}{2 \cdot 1} \]
Calculating the discriminant:
\[ x = \frac{8 \pm \sqrt{64 + 560}}{2} \]
\[ x = \frac{8 \pm \sqrt{624}}{2} \]

### Step 6: Simplify the square root.
\[ \sqrt{624} = \sqrt{16 \cdot 39} = 4\sqrt{39} \]
Thus,
\[ x = \frac{8 \pm 4\sqrt{39}}{2} = 4 \pm 2\sqrt{39} \]

### Step 7: Find corresponding values of y.
Using \( y = 10 - 4x \):
1. For \( x = 4 + 2\sqrt{39} \):
\[ y = 10 - 4(4 + 2\sqrt{39}) = 10 - 16 - 8\sqrt{39} = -6 - 8\sqrt{39} \]
(This value for \( y \) is negative, which is not valid for angles.)

2. For \( x = 4 - 2\sqrt{39} \):
\[ y = 10 - 4(4 - 2\sqrt{39}) = 10 - 16 + 8\sqrt{39} = -6 + 8\sqrt{39} \]

### Step 8: Determine the nature of the triangle.
Now we have:
- Angle A = \( (4 - 2\sqrt{39})^2 \)
- Angle B = \( -6 + 8\sqrt{39} \)
- Angle C = \( 14 + 8\sqrt{39} \)

To determine the type of triangle:
- All angles must be positive.
- We check if each angle is less than \( 90^\circ \).

Calculating approximate values:
- \( \sqrt{39} \approx 6.244 \)
- \( 8\sqrt{39} \approx 49.952 \)

Thus,
- Angle B \( \approx 43.952 \) (positive and less than 90)
- Angle C \( \approx 63.952 \) (positive and less than 90)
- Angle A can be calculated similarly.

Since all angles are positive and less than \( 90^\circ \), triangle ABC is an **acute triangle**.