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 $A B C$ if $\angle A=39.8^{\circ}, a=183.3$, and $b=248.6$. \[ \sin B= \] (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. \[ \angle B_{1}=\square \text { and } \angle B_{2}= \] (round these and all remaining answers to 1 decimal place) Thus, two triangles satisfy the given conditions: triangle $A_{1} B_{1} C_{1}$ and triangle $A_{2} B_{2} C_{2}$. Solve the first triangle: $A_{1} B_{1} C_{1}$ \[ \angle C_{1}=\square \text { and } c_{1}= \] Solve the second triangle: $A_{2} B_{2} C_{2}$ \[ \angle C_{2}=\square \text { and } c_{2}= \] Submit Question

Step 1 ：Use the Law of Sines to find the sine of angle B: \(\sin B = \frac{b \sin A}{a} = \frac{248.6 \sin 39.8^\circ}{183.3} = 0.868146597337479\)

Step 2 ：Find the two possible angles B that have this sine value: \(B_1 = \sin^{-1}(0.868146597337479) = 60.24397128668158^\circ\) and \(B_2 = 180^\circ - B_1 = 119.75602871331841^\circ\)

Step 3 ：Use the sum of angles in a triangle to find the two possible angles C: \(C_1 = 180^\circ - A - B_1 = 79.9560287133184^\circ\) and \(C_2 = 180^\circ - A - B_2 = 20.443971286681574^\circ\)

Step 4 ：Use the Law of Sines again to find the two possible lengths of side c: \(c_1 = \frac{a \sin C_1}{\sin A} = 281.96851352770335\) and \(c_2 = \frac{a \sin C_2}{\sin A} = 100.02205440299642\)

Step 5 ：Final Answer: The two possible angles B are \(\boxed{60.2^\circ}\) and \(\boxed{119.8^\circ}\). The two possible angles C are \(\boxed{80.0^\circ}\) and \(\boxed{20.4^\circ}\). The two possible lengths of side c are \(\boxed{282.0}\) and \(\boxed{100.0}\)