Solve the triangle. \[ \mathrm{a}=8.63 \mathrm{~m}, \mathrm{c}=6.18 \mathrm{~m}, \mathrm{~B}=100.5^{\circ} \] Select the correct choice below and, if necessary, fill in the answer box to complete your choice. (Round all side lengths to the nearest hundredth of a meter. Round all angle measures to the nearest tenth of a degree.) A. There is only 1 possible solution for the triangle. The measurements for the remaining angles \( A \) and \( C \) and side \( b \) are as follows. \( \mathrm{A} \approx \) \( \square \) \( \square^{\circ} \) \( C \approx \) \( \square \) \( { }^{0} \) \( \mathrm{b} \approx \) \( \square \) m B. There are 2 possible solutions for the triangle. The measurements for the solution with the longer side \( b \) are as follows. \( \mathrm{A}= \) \( \square \) \( { }^{\circ} \) \( a \approx \) \( \square \) m \( \mathrm{c} \approx \) \( \square \) m The measurements for the solution with the shorter side b are as follows. \[ \mathrm{A}=\square^{\circ} \] \[ a \approx \] \( \square \) m \( c \approx \) \( \square \) m C. There are no possible solutions for this triangle.

Solve the equation \( \cos(B) = \frac{a^2+c^2-b^2}{2ac} \). Solve the equation by following steps: - step0: Solve for \(B\): \(\cos\left(B\right)=\frac{a^{2}+c^{2}-b^{2}}{2ac}\) - step1: Use the inverse trigonometric function: \(B=\arccos\left(\frac{a^{2}+c^{2}-b^{2}}{2ac}\right)\) - step2: Calculate: \(\begin{align}&B=\arccos\left(\frac{a^{2}+c^{2}-b^{2}}{2ac}\right)\\&B=-\arccos\left(\frac{a^{2}+c^{2}-b^{2}}{2ac}\right)\end{align}\) - step3: Add the period: \(\begin{align}&B=\arccos\left(\frac{a^{2}+c^{2}-b^{2}}{2ac}\right)+2k\pi ,k \in \mathbb{Z}\\&B=-\arccos\left(\frac{a^{2}+c^{2}-b^{2}}{2ac}\right)+2k\pi ,k \in \mathbb{Z}\end{align}\) - step4: Find the union: \(B=\left\{ \begin{array}{l}\arccos\left(\frac{a^{2}+c^{2}-b^{2}}{2ac}\right)+2k\pi \\-\arccos\left(\frac{a^{2}+c^{2}-b^{2}}{2ac}\right)+2k\pi \end{array}\right.,k \in \mathbb{Z}\) The solution for \( B \) is given by the equation \( B = \arccos\left(\frac{a^2+c^2-b^2}{2ac}\right) + 2k\pi \), where \( k \) is an integer. Given that \( a = 8.63 \) m, \( c = 6.18 \) m, and \( B = 100.5^\circ \), we can substitute these values into the equation to find the possible values for \( B \). Substituting the given values into the equation, we have: \[ B = \arccos\left(\frac{8.63^2+6.18^2-b^2}{2 \cdot 8.63 \cdot 6.18}\right) + 2k\pi \] To determine the possible solutions for the triangle, we need to find the values of \( b \) that satisfy the equation. Let's calculate the value of \( b \) using the given values of \( a \) and \( c \). Simplify the expression by following steps: - step0: Solution: \(\arccos\left(\frac{8.63^{2}+6.18^{2}-b^{2}}{2\times 8.63\times 6.18}\right)\) - step1: Convert the expressions: \(\arccos\left(\frac{\left(\frac{863}{100}\right)^{2}+6.18^{2}-b^{2}}{2\times 8.63\times 6.18}\right)\) - step2: Convert the expressions: \(\arccos\left(\frac{\left(\frac{863}{100}\right)^{2}+\left(\frac{309}{50}\right)^{2}-b^{2}}{2\times 8.63\times 6.18}\right)\) - step3: Multiply the terms: \(\arccos\left(\frac{\left(\frac{863}{100}\right)^{2}+\left(\frac{309}{50}\right)^{2}-b^{2}}{106.6668}\right)\) - step4: Calculate: \(\arccos\left(\frac{\frac{863^{2}+618^{2}}{100^{2}}-b^{2}}{106.6668}\right)\) - step5: Simplify the expression: \(\arccos\left(\frac{863^{2}+618^{2}}{1066668}-\frac{2500}{266667}b^{2}\right)\) The value of \( b \) is not directly calculable from the given information. To determine the possible solutions for the triangle, we need to consider the range of values for \( b \) that satisfy the equation. Since the equation involves the arccosine function, the value of \( b \) must be within the range of \( -1 \leq \frac{a^2+c^2-b^2}{2ac} \leq 1 \) to ensure that the arccosine function is defined. Let's calculate the range of values for \( b \) that satisfy the equation. Solve the system of inequalities by following steps: - step0: Solve for \(b\): \(\left\{ \begin{array}{l}-1\leq \frac{8.63^{2}+6.18^{2}-b^{2}}{2\times 8.63\times 6.18}\\\frac{8.63^{2}+6.18^{2}-b^{2}}{2\times 8.63\times 6.18}\leq 1\end{array}\right.\) - step1: Solve the inequality: \(\left\{ \begin{array}{l}-\frac{1481}{100}\leq b\leq \frac{1481}{100}\\b \in \left(-\infty,-\frac{49}{20}\right]\cup \left[\frac{49}{20},+\infty\right)\end{array}\right.\) - step2: Find the intersection: \(b \in \left[-\frac{1481}{100},-\frac{49}{20}\right]\cup \left[\frac{49}{20},\frac{1481}{100}\right]\) The range of values for \( b \) that satisfy the equation is \( b \in \left[-\frac{1481}{100}, -\frac{49}{20}\right] \cup \left[\frac{49}{20}, \frac{1481}{100}\right] \). Since the range of values for \( b \) is not a single value, there are multiple possible solutions for the triangle. Therefore, the correct answer is: B. There are 2 possible solutions for the triangle.