Solve the triangle with \( a=38.37, b=38.15 \), and \( \beta=70^{\circ} \) \( \alpha=\square \) degrees \( \gamma=\square \) degrees \( c=\square \) Round all answers to 2 places after the decimal point. Give angles in degrees (but do not type "degrees" in the answer box). As in the text, side \( a \) is opposite from angle \( \alpha \), side \( b \) is opposite from angle \( \beta \), and side \( c \) is opposite from angle \( \gamma \). If no such triangle exists, then enter DNE in each answer box. If multiple triangles exist, then enter the values of the missing sides and angles for each triangle, separated by commas.

Alright, I need to solve this triangle problem where I'm given two sides and one angle. The sides are \( a = 38.37 \) and \( b = 38.15 \), and the angle opposite side \( b \) is \( \beta = 70^\circ \). I need to find the remaining angle \( \alpha \), the third angle \( \gamma \), and the third side \( c \). All answers should be rounded to two decimal places, and angles should be in degrees without the "degrees" label. First, I recall that in any triangle, the sum of the interior angles is \( 180^\circ \). So, \( \alpha + \beta + \gamma = 180^\circ \). Given \( \beta = 70^\circ \), I can express \( \gamma \) as \( \gamma = 180^\circ - \alpha - 70^\circ = 110^\circ - \alpha \). Next, I can use the Law of Sines to find angle \( \alpha \). The Law of Sines states that \( \frac{a}{\sin \alpha} = \frac{b}{\sin \beta} \). Plugging in the known values: \[ \frac{38.37}{\sin \alpha} = \frac{38.15}{\sin 70^\circ} \] First, I'll calculate \( \sin 70^\circ \). Using a calculator, \( \sin 70^\circ \approx 0.9397 \). So, \[ \frac{38.37}{\sin \alpha} = \frac{38.15}{0.9397} \approx 40.67 \] Now, solving for \( \sin \alpha \): \[ \sin \alpha = \frac{38.37}{40.67} \approx 0.943 \] Taking the inverse sine to find \( \alpha \): \[ \alpha \approx \arcsin(0.943) \approx 70.0^\circ \] Wait a second, that seems a bit off. Let me double-check my calculations. Maybe I made a mistake in the Law of Sines application. Let me recalculate: \[ \frac{38.37}{\sin \alpha} = \frac{38.15}{0.9397} \approx 40.67 \] So, \[ \sin \alpha = \frac{38.37}{40.67} \approx 0.943 \] Taking the arcsin: \[ \alpha \approx \arcsin(0.943) \approx 70.0^\circ \] Hmm, it still comes out to approximately \( 70^\circ \). That seems consistent. But let's verify the angles sum up to \( 180^\circ \): \[ \alpha + \beta + \gamma = 70.0^\circ + 70^\circ + \gamma = 180^\circ \] So, \[ \gamma = 180^\circ - 140^\circ = 40.0^\circ \] Now, to find side \( c \), I can use the Law of Sines again: \[ \frac{c}{\sin \gamma} = \frac{b}{\sin \beta} \] Plugging in the known values: \[ \frac{c}{\sin 40.0^\circ} = \frac{38.15}{0.9397} \approx 40.67 \] Calculating \( \sin 40.0^\circ \): \[ \sin 40.0^\circ \approx 0.6428 \] So, \[ c = 40.67 \times 0.6428 \approx 26.00 \] Let me verify these calculations to ensure accuracy. Using a calculator for \( \sin 70^\circ \) and \( \sin 40^\circ \) seems correct, and the arithmetic checks out. Therefore, the triangle can be solved with these values.