If `sinA=sin^2B and 2cos^2A=3cos^2B` then the triangle ABC is

Options

A. right angled
B. obtuse angled
C. ospsceles
D. equilateral

Explanation

To determine the type of triangle ABC given the conditions \( \sin A = \sin^2 B \) and \( 2 \cos^2 A = 3 \cos^2 B \), we can follow these steps:

### Step 1: Rewrite the second equation
We start with the equation:
\[ 2 \cos^2 A = 3 \cos^2 B \]
Using the identity \( \cos^2 \theta = 1 - \sin^2 \theta \), we can rewrite this as:
\[ 2(1 - \sin^2 A) = 3(1 - \sin^2 B) \]

### Step 2: Expand and rearrange
Expanding both sides gives:
\[ 2 - 2 \sin^2 A = 3 - 3 \sin^2 B \]
Rearranging this, we get:
\[ 2 - 3 = 2 \sin^2 A - 3 \sin^2 B \]
which simplifies to:
\[ -1 = 2 \sin^2 A - 3 \sin^2 B \]
or
\[ 2 \sin^2 A - 3 \sin^2 B + 1 = 0 \]

### Step 3: Substitute \( \sin^2 B \)
From the first equation, we know \( \sin A = \sin^2 B \). Therefore, we can express \( \sin^2 B \) in terms of \( \sin A \):
\[ \sin^2 B = \sin A \]
Substituting this into our equation gives:
\[ 2 \sin^2 A - 3 \sin A + 1 = 0 \]

### Step 4: Solve the quadratic equation
Now we will solve the quadratic equation:
\[ 2 \sin^2 A - 3 \sin A + 1 = 0 \]
Using the quadratic formula \( \sin A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 2 \), \( b = -3 \), and \( c = 1 \).
- The discriminant is:
\[ b^2 - 4ac = (-3)^2 - 4 \cdot 2 \cdot 1 = 9 - 8 = 1 \]
Thus, we have:
\[ \sin A = \frac{3 \pm 1}{4} \]
Calculating the two possible values:
1. \( \sin A = \frac{4}{4} = 1 \) (which gives \( A = 90^\circ \))
2. \( \sin A = \frac{2}{4} = \frac{1}{2} \) (which gives \( A = 30^\circ \))

### Step 5: Determine the corresponding angles
1. If \( A = 90^\circ \), then \( \sin A = 1 \) implies \( \sin^2 B = 1 \) which means \( B = 90^\circ \). This case is not possible in a triangle.
2. If \( A = 30^\circ \), then \( \sin^2 B = \frac{1}{2} \) implies \( B = 45^\circ \).

### Step 6: Find angle C
Using the triangle angle sum property:
\[ C = 180^\circ - A - B = 180^\circ - 30^\circ - 45^\circ = 105^\circ \]

### Conclusion: Type of triangle
Since one angle \( C = 105^\circ \) is greater than \( 90^\circ \), triangle ABC is an **obtuse triangle**.