To determine how many distinct triangles can be formed given [tex]\( m \angle A = 75^\circ \)[/tex], [tex]\( a = 2 \)[/tex], and [tex]\( b = 3 \)[/tex] using the law of sines, we must follow these steps:
1. Convert the given angle [tex]\( A \)[/tex] to radians:
[tex]\[
A = 75^\circ
\][/tex]
2. Using the law of sines, calculate [tex]\(\sin(A)/a\)[/tex]:
[tex]\[
\sin(A) / a = \sin(75^\circ) / 2
\][/tex]
[tex]\[
\sin(75^\circ) = \sin(45^\circ + 30^\circ) = \sin(45^\circ)\cos(30^\circ) + \cos(45^\circ)\sin(30^\circ)
\][/tex]
Using known values:
[tex]\[
\sin(45^\circ) = \cos(45^\circ) = \frac{\sqrt{2}}{2}, \quad \cos(30^\circ) = \frac{\sqrt{3}}{2}, \quad \sin(30^\circ) = \frac{1}{2}
\][/tex]
[tex]\[
\sin(75^\circ) = \left( \frac{\sqrt{2}}{2} \right) \left( \frac{\sqrt{3}}{2} \right) + \left( \frac{\sqrt{2}}{2} \right) \left( \frac{1}{2} \right)
\][/tex]
[tex]\[
\sin(75^\circ) = \frac{\sqrt{2}\sqrt{3}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}
\][/tex]
Then,
[tex]\[
\sin(A) / a = \frac{ \frac{\sqrt{6} + \sqrt{2}}{4} }{2} = \frac{\sqrt{6} + \sqrt{2}}{8}
\][/tex]
3. Determine [tex]\(\sin(B)\)[/tex] using the law of sines:
[tex]\[
\sin(B) = b \cdot (\sin(A) / a) = 3 \cdot \frac{\sqrt{6} + \sqrt{2}}{8}
\][/tex]
[tex]\[
\sin(B) = \frac{3(\sqrt{6} + \sqrt{2})}{8}
\][/tex]
4. Check if [tex]\(\sin(B)\)[/tex] is within the valid range for sine function [tex]\([-1, 1]\)[/tex]:
[tex]\[
\sin(B) = \frac{3(\sqrt{6} + \sqrt{2})}{8}
\][/tex]
Evaluating this value,
[tex]\[
\sqrt{6} \approx 2.45, \quad \sqrt{2} \approx 1.41
\][/tex]
[tex]\[
\sin(B) \approx \frac{3(2.45 + 1.41)}{8} = \frac{3 \times 3.86}{8} = \frac{11.58}{8} \approx 1.4475
\][/tex]
Since [tex]\(\sin(B)\)[/tex] is approximately 1.4475, it is greater than 1.
Since [tex]\(\sin(B) > 1\)[/tex], it implies that no real angle [tex]\(B\)[/tex] can satisfy these conditions within the triangle, and therefore, no triangle can be formed.
Thus, the answer is:
No triangles can be formed.
