Question

remember, even if you enter an answer rounded to a set number of decimal places, if you use that number in a future calculation, you should use all of the decimal places reported on your calculator!
solve triangle abc if \\( \angle a = 35.9^{\circ} \\), \\( a = 186.2 \\), and \\( b = 245.7 \\).
\\( \sin b = \square \\)
(round answer to 5 decimal places)
there are two possible angles b between \\( 0^{\circ} \\) and \\( 180^{\circ} \\) with this value for sine. find the two angles, and report them so that \\( \angle b_1 \\) is the acute angle.
\\( \angle b_1 = \square^{\circ} \\) and \\( \angle b_2 = \square^{\circ} \\)
(round these and all remaining answers to 1 decimal place)
thus, two triangles satisfy the given conditions: triangle \\( a_1b_1c_1 \\) and triangle \\( a_2b_2c_2 \\).
solve the first triangle: \\( a_1b_1c_1 \\)
\\( \angle c_1 = \square^{\circ} \\) and \\( c_1 = \square \\)
solve the second triangle: \\( a_2b_2c_2 \\)
\\( \angle c_2 = \square^{\circ} \\) and \\( c_2 = \square \\)

Explanation:

Step1: Use the Law of Sines

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}\). We are given \(A = 35.9^{\circ}\), \(a = 186.2\), and \(b = 245.7\). Rearranging for \(\sin B\), we get \(\sin B=\frac{b\sin A}{a}\).
Substituting the values: \(\sin B=\frac{245.7\times\sin(35.9^{\circ})}{186.2}\)
First, find \(\sin(35.9^{\circ})\approx0.58682\)
Then \(\sin B=\frac{245.7\times0.58682}{186.2}\approx\frac{144.17267}{186.2}\approx0.77439\)

Step2: Find the two angles \(B\)

Since \(\sin B = 0.77439\), the acute angle \(B_1=\sin^{- 1}(0.77439)\approx50.8^{\circ}\)
The obtuse angle \(B_2 = 180^{\circ}-B_1=180^{\circ}-50.8^{\circ}=129.2^{\circ}\)

Step3: Solve for \(C_1\) and \(c_1\) in triangle \(A_1B_1C_1\)

For \(C_1\): Using the angle - sum property of a triangle (\(A + B+C=180^{\circ}\)), \(C_1=180^{\circ}-A - B_1\)
\(C_1=180^{\circ}-35.9^{\circ}-50.8^{\circ}=93.3^{\circ}\)
Using the Law of Sines \(\frac{c_1}{\sin C_1}=\frac{a}{\sin A}\), so \(c_1=\frac{a\sin C_1}{\sin A}\)
\(\sin C_1=\sin(93.3^{\circ})\approx0.9985\), \(\sin A=\sin(35.9^{\circ})\approx0.58682\)
\(c_1=\frac{186.2\times0.9985}{0.58682}\approx\frac{185.92}{0.58682}\approx316.8\)

Step4: Solve for \(C_2\) and \(c_2\) in triangle \(A_2B_2C_2\)

For \(C_2\): \(C_2=180^{\circ}-A - B_2\)
\(C_2=180^{\circ}-35.9^{\circ}-129.2^{\circ}=14.9^{\circ}\)
Using the Law of Sines \(\frac{c_2}{\sin C_2}=\frac{a}{\sin A}\), so \(c_2=\frac{a\sin C_2}{\sin A}\)
\(\sin C_2=\sin(14.9^{\circ})\approx0.2571\)
\(c_2=\frac{186.2\times0.2571}{0.58682}\approx\frac{47.86}{0.58682}\approx81.6\)

Answer:

\(\sin B\approx0.77439\)
\(\angle B_1\approx50.8^{\circ}\), \(\angle B_2\approx129.2^{\circ}\)
\(\angle C_1\approx93.3^{\circ}\), \(c_1\approx316.8\)
\(\angle C_2\approx14.9^{\circ}\), \(c_2\approx81.6\)