To solve for side \( b \) in triangle \( ABC \) using the Law of Sines, we first need the relevant angles and sides. We know:
- \( \angle A = 45^\circ \)
- \( \angle B = 65^\circ \)
- \( a = 15.05 \)
First, we can find angle \( C \) using the fact that the sum of the angles in a triangle is \( 180^\circ \):
\[ \angle C = 180^\circ - \angle A - \angle B \] \[ \angle C = 180^\circ - 45^\circ - 65^\circ = 70^\circ \]
Now we can apply the Law of Sines, which states:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} \]
Substituting our known values:
\[ \frac{15.05}{\sin 45^\circ} = \frac{b}{\sin 65^\circ} \]
Next, we calculate \( \sin 45^\circ \) and \( \sin 65^\circ \):
\[ \sin 45^\circ = \frac{\sqrt{2}}{2} \approx 0.7071 \] \[ \sin 65^\circ \approx 0.9063 \]
Now we can plug these values into our equation:
\[ \frac{15.05}{0.7071} = \frac{b}{0.9063} \]
Simplifying the left-hand side:
\[ 15.05 \div 0.7071 \approx 21.24 \]
Now we have:
\[ 21.24 = \frac{b}{0.9063} \]
To find \( b \), we solve for \( b \):
\[ b = 21.24 \times 0.9063 \approx 19.29 \]
Rounding to the nearest tenth, we find:
\[ b \approx 19.3 \]
Thus, the length of side \( b \) is approximately 19.3.
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