Question

this problem asks you to
edo\ example #4 in this section with different numbers. read this example carefully before attempting this problem.
solve triangle ( abc ) if ( angle a = 43.1^{circ}, a = 187.6 ), and ( b = 247.1 ).
( sin b=) (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 = 64.2 ) and ( angle b_2 = 115.8 )
(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 = 72.7 ) and ( c_1 = 262.2 )
solve the second triangle: ( a_2b_2c_2 )
( angle c_2 = 21.1 ) and ( c_2 = 99.03 )

Explanation:

Step1: Use the Law of Sines

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}\). We can rearrange this formula to solve for \(\sin B\): \(\sin B=\frac{b\sin A}{a}\).

Step2: Substitute the given values

Given \(A = 43.1^{\circ}\), \(a = 187.6\), and \(b = 247.1\). First, find \(\sin A=\sin(43.1^{\circ})\approx0.6830\). Then \(\sin B=\frac{247.1\times0.6830}{187.6}\).

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Answer:

\(\sin B\approx0.89914\)