Law of sines: [tex]\frac{\sin (A)}{a}=\frac{\sin (B)}{b}=\frac{\sin (C)}{c}[/tex]

How many distinct triangles can be formed for which [tex]m \angle A=75^{\circ}, a=2[/tex], and [tex]b=3[/tex]?

A. No triangles can be formed.
B. One triangle can be formed where [tex]\angle B[/tex] is about [tex]15^{\circ}[/tex].
C. One triangle can be formed where [tex]\angle B[/tex] is about [tex]40^{\circ}[/tex].
D. Two triangles can be formed where [tex]\angle B[/tex] is [tex]40^{\circ}[/tex] or [tex]140^{\circ}[/tex].



Answer :

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.