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 ∠a = 43.1°, a = 185.5, and b = 243.7.

sin b =
(round answer to 5 decimal places)
there are two possible angles b between 0° and 180° with this value for sine. find the two angles, and report them so that ∠b₁ is the acute angle.
∠b₁ = ° and ∠b₂ = °
(round these and all remaining answers to 1 decimal place)
thus, two triangles satisfy the given conditions: triangle a₁b₁c₁ and triangle a₂b₂c₂.

solve the first triangle: a₁b₁c₁
∠c₁ = ° and c₁ =

solve the second triangle: a₂b₂c₂
∠c₂ = ° and c₂ =

Explanation:

Step1: Use the Law of Sines

The Law of Sines is given by \(\frac{a}{\sin A}=\frac{b}{\sin B}\).
We know that \(a = 185.5\), \(A=43.1^{\circ}\), and \(b = 243.7\).
Substitute these values into the Law of Sines formula: \(\sin B=\frac{b\sin A}{a}\).

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Step2: Find the acute angle \(B_1\)

If \(\sin B = 0.89729\), then \(B_1=\sin^{- 1}(0.89729)\approx63.9^{\circ}\)

Step3: Find the obtuse angle \(B_2\)

Since \(\sin\theta=\sin(180^{\circ}-\theta)\), \(B_2 = 180^{\circ}-B_1\)
\(B_2=180^{\circ}- 63.9^{\circ}=116.1^{\circ}\)

Step4: Find \(C_1\) for the first triangle

We know that \(A + B_1+C_1=180^{\circ}\), so \(C_1=180^{\circ}-(A + B_1)\)
\(C_1=180^{\circ}-(43.1^{\circ}+63.9^{\circ})=73.0^{\circ}\)

Step5: Find \(c_1\) for the first triangle

Using the Law of Sines \(\frac{c_1}{\sin C_1}=\frac{a}{\sin A}\)
\(c_1=\frac{a\sin C_1}{\sin A}=\frac{185.5\times\sin(73.0^{\circ})}{\sin(43.1^{\circ})}\)

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Step6: Find \(C_2\) for the second triangle

We know that \(A + B_2+C_2=180^{\circ}\), so \(C_2=180^{\circ}-(A + B_2)\)
\(C_2=180^{\circ}-(43.1^{\circ}+116.1^{\circ})=20.8^{\circ}\)

Step7: Find \(c_2\) for the second triangle

Using the Law of Sines \(\frac{c_2}{\sin C_2}=\frac{a}{\sin A}\)
\(c_2=\frac{a\sin C_2}{\sin A}=\frac{185.5\times\sin(20.8^{\circ})}{\sin(43.1^{\circ})}\)

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Answer:

\(\sin B\approx0.89729\), \(\angle B_1 = 63.9^{\circ}\), \(\angle B_2=116.1^{\circ}\), \(\angle C_1 = 73.0^{\circ}\), \(c_1\approx259.5\), \(\angle C_2=20.8^{\circ}\), \(c_2\approx96.4\)