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Solve the triangle. \( \mathrm{a}=96.0 \) in., \( \mathrm{b}=60.0 \mathrm{in} ., \mathrm{B}=25.0^{\circ} \) Select the correct choice below and fill in the answer boxes within the choice. Round to the nearest tenth as needed.) A. There is only 1 possible solution for the triangle. The measurements for the remaining angles A and C and side c are as follows. \( \mathrm{A} \approx \square^{\circ} \) \( \mathrm{C} \approx \square^{\circ} \) B . There are 2 possible solutions for the triangle. The measurements for the solution with the longer side c are as follows. \( \mathrm{A} \approx \square^{\circ} \) \( \mathrm{C} \approx \square^{\circ} \) The measurements for the solution with the shorter side c are as follows. \( \mathrm{A} \approx \square^{\circ} \) \( \mathrm{C} \approx \square^{\circ} \) C. There are no possible solutions for this triangle. \( \mathrm{C} \approx \square^{\circ} \) in. in.

Ask by Hall Willis. in the United States
Feb 03,2025

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Answer

There are 2 possible solutions for the triangle. \( \mathrm{A} \approx 42.5^{\circ} \), \( \mathrm{C} \approx 112.5^{\circ} \) and \( \mathrm{A} \approx 137.5^{\circ} \), \( \mathrm{C} \approx 17.5^{\circ} \).

Solution

To solve the given triangle with \( a = 96.0 \) inches, \( b = 60.0 \) inches, and \( \angle B = 25.0^\circ \), we'll use the **Law of Sines**. This problem falls under the SSA (Side-Side-Angle) scenario, which can result in **0, 1, or 2** possible solutions. ### Step 1: Apply the Law of Sines \[ \frac{\sin A}{a} = \frac{\sin B}{b} \] \[ \sin A = \frac{a \cdot \sin B}{b} = \frac{96.0 \cdot \sin 25.0^\circ}{60.0} \approx \frac{96.0 \cdot 0.4226}{60.0} \approx 0.6762 \] ### Step 2: Determine Possible Angles for \( \angle A \) Since \( \sin A \approx 0.6762 \), there are **two possible angles**: 1. **Primary Solution:** \[ A_1 = \arcsin(0.6762) \approx 42.5^\circ \] 2. **Secondary Solution:** \[ A_2 = 180^\circ - 42.5^\circ = 137.5^\circ \] ### Step 3: Calculate Remaining Angles - **For the Primary Solution (\( A \approx 42.5^\circ \)):** \[ C = 180^\circ - B - A_1 = 180^\circ - 25.0^\circ - 42.5^\circ = 112.5^\circ \] - **For the Secondary Solution (\( A \approx 137.5^\circ \)):** \[ C = 180^\circ - B - A_2 = 180^\circ - 25.0^\circ - 137.5^\circ = 17.5^\circ \] ### Step 4: Determine Side \( c \) for Both Solutions - **Primary Solution (Longer side \( c \)):** \[ c = \frac{a \cdot \sin C}{\sin A} = \frac{96.0 \cdot \sin 112.5^\circ}{\sin 42.5^\circ} \approx 131.1 \text{ inches} \] - **Secondary Solution (Shorter side \( c \)):** \[ c = \frac{a \cdot \sin C}{\sin A} = \frac{96.0 \cdot \sin 17.5^\circ}{\sin 137.5^\circ} \approx 42.7 \text{ inches} \] ### Conclusion There are **two possible solutions** for the triangle. **Answer:** **B. There are 2 possible solutions for the triangle.** The measurements for the solution with the longer side \( c \) are as follows. \( \mathrm{A} \approx 42.5^{\circ} \) \( \mathrm{C} \approx 112.5^{\circ} \) The measurements for the solution with the shorter side \( c \) are as follows. \( \mathrm{A} \approx 137.5^{\circ} \) \( \mathrm{C} \approx 17.5^{\circ} \)

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The Deep Dive

To solve the triangle given \( a = 96.0 \) in., \( b = 60.0 \) in., and \( B = 25.0^\circ \), we can use the Law of Sines and some trigonometry. First, we find angle A using the formula: \[ \frac{a}{\sin A} = \frac{b}{\sin B} \implies \sin A = \frac{a \cdot \sin B}{b} = \frac{96 \cdot \sin(25)}{60} \] Calculating \( \sin A \), we find \( A \approx 46.6^\circ \). Next, we can find angle C: \[ C = 180^\circ - A - B = 180^\circ - 46.6^\circ - 25^\circ \approx 108.4^\circ \] Now, to find side c, we use the Law of Sines again: \[ \frac{c}{\sin C} = \frac{a}{\sin A} \implies c = \frac{a \cdot \sin C}{\sin A} \] Calculating \( c \), we find \( c \approx 113.4 \) in. So the answers are: A. There is only 1 possible solution for the triangle. The measurements for the remaining angles A and C and side c are as follows: \( A \approx 46.6^\circ \) \( C \approx 108.4^\circ \) Let’s take a moment to understand the magic of triangles! Did you know that the very word "triangle" comes from the Latin "triangulus," which means "three angles"? Throughout history, triangles have been a symbol of strength and unity, evident in everything from the pyramids of Egypt to the shapes of bridges. In the real world, triangles are everywhere! Architects and engineers rely on triangles' inherent strength in structures. You might find them in roof trusses, bridges, and even in everyday objects like the wedge of a doorstop. Understanding how to solve triangles is essential for everything from building skyscrapers to designing tiny, intricate electronics!

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