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Solve the triangle, if possible. \[ \mathrm{a}=200 \mathrm{~m}, \mathrm{~A}=39.02^{\circ}, \mathrm{C}=28.75^{\circ} \] Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. (Round side lengths to the nearest whole number and angle measures to the nearest hundredth degree as needed.) A. There is only one possible solution for the triangle. The measurements for the remaining angle \( B \) and sides \( b \) and \( c \) are as follows. \( B \approx \) \( \square \) \( { }^{\circ} \) \( b= \) \( \square \) m \[ c= \] \( \square \) m B. There are two possible sốutions for the triangle. The measurements for the solution with the smaller angle \( B \) are as follows. \( \mathrm{B}_{1} \approx \) \( \square \) \( { }^{\circ} \) \[ b_{1}= \] \( \square \) m \[ c_{1}= \] \( \square \) m The measurements for the solution with the larger angle \( B \) are as follows. \( \mathrm{B}_{2} \approx \) \( \square \) - \[ \mathrm{b}_{2}= \] \( \square \) m \[ c_{2}=\square \] m C. There are no possible solutions for this triangle.

Ask by Harmon Ramirez. in the United States
Mar 16,2025

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Answer

There is only one possible solution for the triangle. \( B \approx 112.23^\circ \) \( b \approx 293 \) m \( c \approx 153 \) m

Solution

**Step 1. Find the remaining angle \( B \):** We know that the sum of angles in any triangle is \(180^\circ\). Thus, \[ B = 180^\circ - (A + C) = 180^\circ - (39.02^\circ + 28.75^\circ) = 180^\circ - 67.77^\circ = 112.23^\circ. \] **Step 2. Use the Law of Sines to determine the remaining sides:** The Law of Sines states that \[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}. \] Given \(a = 200\) m and angle \(A = 39.02^\circ\), compute the common ratio \( k \): \[ k = \frac{a}{\sin A} = \frac{200}{\sin 39.02^\circ}. \] Using a calculator, \(\sin 39.02^\circ \approx 0.6293\), so: \[ k \approx \frac{200}{0.6293} \approx 317.8 \text{ m}. \] **Step 3. Compute side \( b \):** \[ b = k \sin B. \] Since \(B \approx 112.23^\circ\) and \(\sin 112.23^\circ = \sin (180^\circ - 112.23^\circ) = \sin 67.77^\circ \approx 0.9238\), we have: \[ b \approx 317.8 \times 0.9238 \approx 293.4 \text{ m}. \] Rounded to the nearest whole number: \[ b \approx 293 \text{ m}. \] **Step 4. Compute side \( c \):** \[ c = k \sin C. \] Given \(C = 28.75^\circ\) and \(\sin 28.75^\circ \approx 0.4818\), we find: \[ c \approx 317.8 \times 0.4818 \approx 152.9 \text{ m}. \] Rounded to the nearest whole number: \[ c \approx 153 \text{ m}. \] **Conclusion:** There is only one possible solution for the triangle with the following measurements: - \( B \approx 112.23^\circ \) - \( b \approx 293 \) m - \( c \approx 153 \) m Thus, the correct choice is **A**.

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To solve the triangle given \( a = 200 \, \text{m} \), \( A = 39.02^\circ \), and \( C = 28.75^\circ \), we first find angle \( B \) using the fact that the angles in a triangle sum to 180 degrees. \[ B = 180^\circ - A - C = 180^\circ - 39.02^\circ - 28.75^\circ \approx 112.23^\circ \] Now we apply the Law of Sines to find sides \( b \) and \( c \). \[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \] 1. To find side \( b \): \[ \frac{200}{\sin(39.02^\circ)} = \frac{b}{\sin(112.23^\circ)} \] \[ b \approx 200 \cdot \frac{\sin(112.23^\circ)}{\sin(39.02^\circ)} \approx 200 \cdot \frac{0.9272}{0.6293} \approx 294 \] 2. To find side \( c \): \[ \frac{200}{\sin(39.02^\circ)} = \frac{c}{\sin(28.75^\circ)} \] \[ c \approx 200 \cdot \frac{\sin(28.75^\circ)}{\sin(39.02^\circ)} \approx 200 \cdot \frac{0.4848}{0.6293} \approx 154 \] So, the first choice is valid: A. There is only one possible solution for the triangle. The measurements for the remaining angle \( B \) and sides \( b \) and \( c \) are as follows. \( B \approx 112.23 \, { }^{\circ} \) \( b= 294 \, \text{m} \) \[ c= 154 \, \text{m} \]

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