Use the Law of Sines to solve the triangle. \( \mathrm{B}=52^{\circ}, \mathrm{C}=16^{\circ}, \mathrm{b}=42 \) \( \mathrm{~A}=\square^{\circ} \) (Round to the nearest degree as needed.)

Alright, I need to solve this triangle using the Law of Sines. The given information is angle B = 52 degrees, angle C = 16 degrees, and side b = 42 units. I need to find angle A, which is currently represented as a square. Let's break this down step by step. First, I recall that the sum of the angles in any triangle is always 180 degrees. So, if I know two of the angles, I can find the third one. Given that angle B is 52 degrees and angle C is 16 degrees, I can calculate angle A as follows: Angle A = 180 degrees - (Angle B + Angle C) = 180 - (52 + 16) = 180 - 68 = 112 degrees. Wait a second, that seems straightforward, but the question specifically asks to use the Law of Sines to solve the triangle. Maybe I should double-check if the Law of Sines is necessary here or if it's just an extra step. The Law of Sines states that in any triangle, the ratio of the length of a side to the sine of its opposite angle is constant. Mathematically, it's expressed as: \[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \] Given that I already have two angles and one side, I can use the Law of Sines to find the other sides. Let's denote the sides opposite to angles A, B, and C as a, b, and c respectively. Here, side b is given as 42 units, and angle B is 52 degrees. Using the Law of Sines: \[ \frac{a}{\sin A} = \frac{42}{\sin 52^\circ} \] But I need to find angle A first, which I've already calculated as 112 degrees. So, plugging that in: \[ \frac{a}{\sin 112^\circ} = \frac{42}{\sin 52^\circ} \] Calculating the sine values: \[ \sin 112^\circ \approx 0.9272 \] \[ \sin 52^\circ \approx 0.7880 \] Now, solving for side a: \[ a = \frac{42 \times \sin 112^\circ}{\sin 52^\circ} = \frac{42 \times 0.9272}{0.7880} \approx \frac{38.9424}{0.7880} \approx 49.3 \] So, side a is approximately 49.3 units. Similarly, I can find side c using the Law of Sines: \[ \frac{c}{\sin C} = \frac{42}{\sin 52^\circ} \] Given that angle C is 16 degrees: \[ \sin 16^\circ \approx 0.2756 \] So, \[ c = \frac{42 \times \sin 16^\circ}{\sin 52^\circ} = \frac{42 \times 0.2756}{0.7880} \approx \frac{11.5752}{0.7880} \approx 14.6 \] Therefore, side c is approximately 14.6 units. To summarize, the triangle has the following measurements: - Angle A = 112 degrees - Angle B = 52 degrees - Angle C = 16 degrees - Side a ≈ 49.3 units - Side b = 42 units - Side c ≈ 14.6 units Everything checks out, and the calculations seem consistent. I used the Law of Sines effectively to find the missing side lengths after determining angle A.