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 = 189.4, and b = 242.8.

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 states that \(\frac{a}{\sin A}=\frac{b}{\sin B}\). We know \(a = 189.4\), \(A=43.1^{\circ}\), and \(b = 242.8\). Rearranging for \(\sin B\), we get \(\sin B=\frac{b\sin A}{a}\).
Substitute the values: \(\sin B=\frac{242.8\times\sin(43.1^{\circ})}{189.4}\).
Using a calculator, \(\sin(43.1^{\circ})\approx0.6830\). Then \(\sin B=\frac{242.8\times0.6830}{189.4}\approx0.8730\).

Step2: Find the acute angle \(B_1\)

Since \(\sin B = 0.8730\), \(B_1=\sin^{- 1}(0.8730)\approx60.8^{\circ}\) (using a calculator's inverse - sine function).

Step3: Find the obtuse angle \(B_2\)

We know that \(\sin\theta=\sin(180^{\circ}-\theta)\). So \(B_2 = 180^{\circ}-B_1=180^{\circ}-60.8^{\circ}=119.2^{\circ}\).

Step4: Find \(C_1\) in triangle \(A_1B_1C_1\)

In a triangle, \(A + B + C=180^{\circ}\). For \(A = 43.1^{\circ}\) and \(B_1 = 60.8^{\circ}\), \(C_1=180^{\circ}-(43.1^{\circ}+60.8^{\circ})=76.1^{\circ}\).
Using the Law of Sines \(\frac{c_1}{\sin C_1}=\frac{a}{\sin A}\), so \(c_1=\frac{a\sin C_1}{\sin A}\).
Substitute \(a = 189.4\), \(\sin A\approx0.6830\), \(\sin C_1=\sin(76.1^{\circ})\approx0.9703\). Then \(c_1=\frac{189.4\times0.9703}{0.6830}\approx269.3\).

Step5: Find \(C_2\) in triangle \(A_2B_2C_2\)

For \(A = 43.1^{\circ}\) and \(B_2 = 119.2^{\circ}\), \(C_2=180^{\circ}-(43.1^{\circ}+119.2^{\circ})=17.7^{\circ}\).
Using the Law of Sines \(\frac{c_2}{\sin C_2}=\frac{a}{\sin A}\), so \(c_2=\frac{a\sin C_2}{\sin A}\).
Substitute \(a = 189.4\), \(\sin A\approx0.6830\), \(\sin C_2=\sin(17.7^{\circ})\approx0.3040\). Then \(c_2=\frac{189.4\times0.3040}{0.6830}\approx84.3\).

Answer:

\(\sin B = 0.8730\)
\(\angle B_1 = 60.8^{\circ}\), \(\angle B_2 = 119.2^{\circ}\)
\(\angle C_1 = 76.1^{\circ}\), \(c_1 = 269.3\)
\(\angle C_2 = 17.7^{\circ}\), \(c_2 = 84.3\)