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 = 181.7, and b = 246.1.
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 to find \(\sin B\)

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}\).
We are given \(a = 181.7\), \(A=43.1^{\circ}\), and \(b = 246.1\).
Substitute the values into the formula: \(\sin B=\frac{b\sin A}{a}\)
\(\sin B=\frac{246.1\times\sin(43.1^{\circ})}{181.7}\)
First, calculate \(\sin(43.1^{\circ})\approx0.6830\)
Then \(\sin B=\frac{246.1\times0.6830}{181.7}=\frac{168.0863}{181.7}\approx0.92507\)

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

Since \(\sin B = 0.92507\), and \(B\in(0^{\circ},180^{\circ})\)
\(B_1=\sin^{- 1}(0.92507)\approx67.8^{\circ}\)
\(B_2 = 180^{\circ}-B_1=180^{\circ}-67.8^{\circ}=112.2^{\circ}\)

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

For \(\angle C_1\), use the angle - sum property of a triangle \(\angle A+\angle B+\angle C = 180^{\circ}\)
\(\angle C_1=180^{\circ}-\angle A - \angle B_1=180^{\circ}-43.1^{\circ}-67.8^{\circ}=69.1^{\circ}\)
Again, use the Law of Sines \(\frac{a}{\sin A}=\frac{c_1}{\sin C_1}\)
\(c_1=\frac{a\sin C_1}{\sin A}=\frac{181.7\times\sin(69.1^{\circ})}{\sin(43.1^{\circ})}\)
Since \(\sin(69.1^{\circ})\approx0.934\) and \(\sin(43.1^{\circ})\approx0.683\)
\(c_1=\frac{181.7\times0.934}{0.683}=\frac{169.7078}{0.683}\approx248.5\)

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

For \(\angle C_2\), use \(\angle A+\angle B+\angle C = 180^{\circ}\)
\(\angle C_2=180^{\circ}-\angle A-\angle B_2=180^{\circ}-43.1^{\circ}-112.2^{\circ}=24.7^{\circ}\)
Use the Law of Sines \(\frac{a}{\sin A}=\frac{c_2}{\sin C_2}\)
\(c_2=\frac{a\sin C_2}{\sin A}=\frac{181.7\times\sin(24.7^{\circ})}{\sin(43.1^{\circ})}\)
Since \(\sin(24.7^{\circ})\approx0.418\) and \(\sin(43.1^{\circ})\approx0.683\)
\(c_2=\frac{181.7\times0.418}{0.683}=\frac{75.9506}{0.683}\approx111.2\)

Answer:

\(\sin B = 0.92507\)
\(\angle B_1 = 67.8^{\circ}\), \(\angle B_2=112.2^{\circ}\)
\(\angle C_1 = 69.1^{\circ}\), \(c_1 = 248.5\)
\(\angle C_2=24.7^{\circ}\), \(c_2 = 111.2\)