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 = 43.1^circ ), ( a = 184.7 ), and ( b = 241.5 ).
( 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 to find \(\sin B\)

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}\).
Substitute \(a = 184.7\), \(A=43.1^{\circ}\), and \(b = 241.5\) into the formula:
\(\sin B=\frac{b\sin A}{a}=\frac{241.5\times\sin(43.1^{\circ})}{184.7}\)
Using a calculator, \(\sin(43.1^{\circ})\approx0.6830\)
\(\sin B=\frac{241.5\times0.6830}{184.7}=\frac{164.9445}{184.7}\approx0.89303\)

Step2: Find the two possible angles \(B\)

If \(\sin B = 0.89303\), then \(B_1=\sin^{- 1}(0.89303)\approx63.3^{\circ}\)
\(B_2 = 180^{\circ}-B_1=180^{\circ}-63.3^{\circ}=116.7^{\circ}\)

Step3: Find \(\angle C_1\) and \(c_1\) for triangle \(A_1B_1C_1\)

For triangle \(A_1B_1C_1\), \(\angle C_1=180^{\circ}-\angle A-\angle B_1\)
\(\angle C_1=180^{\circ}-43.1^{\circ}-63.3^{\circ}=73.6^{\circ}\)
Using the Law of Sines \(\frac{c_1}{\sin C_1}=\frac{a}{\sin A}\), so \(c_1=\frac{a\sin C_1}{\sin A}\)
\(c_1=\frac{184.7\times\sin(73.6^{\circ})}{\sin(43.1^{\circ})}\)
\(\sin(73.6^{\circ})\approx0.9590\), \(\sin(43.1^{\circ})\approx0.6830\)
\(c_1=\frac{184.7\times0.9590}{0.6830}=\frac{177.1273}{0.6830}\approx259.3\)

Step4: Find \(\angle C_2\) and \(c_2\) for triangle \(A_2B_2C_2\)

For triangle \(A_2B_2C_2\), \(\angle C_2=180^{\circ}-\angle A-\angle B_2\)
\(\angle C_2=180^{\circ}-43.1^{\circ}-116.7^{\circ}=20.2^{\circ}\)
Using the Law of Sines \(\frac{c_2}{\sin C_2}=\frac{a}{\sin A}\), so \(c_2=\frac{a\sin C_2}{\sin A}\)
\(\sin(20.2^{\circ})\approx0.3450\), \(\sin(43.1^{\circ})\approx0.6830\)
\(c_2=\frac{184.7\times0.3450}{0.6830}=\frac{63.7215}{0.6830}\approx93.3\)

Answer:

\(\sin B = 0.89303\)
\(\angle B_1 = 63.3^{\circ}\), \(\angle B_2=116.7^{\circ}\)
\(\angle C_1 = 73.6^{\circ}\), \(c_1 = 259.3\)
\(\angle C_2 = 20.2^{\circ}\), \(c_2 = 93.3\)