Question

current attempt in progress
solve the triangle.
20°
14 in.
a
130°
α
c
o α = 30°, a = 6.3 in., c = 9.1 in.
o α = 30°, a = 9.1 in., c = 6.3 in.
o α = 40°, a = 7.1 in., c = 8.6 in.
o α = 40°, a = 8.6 in., c = 7.1 in.
o no triangle with the given measurements.

Explanation:

Step1: Find the angle \(\alpha\)

The sum of angles in a triangle is \(180^{\circ}\). So \(\alpha=180^{\circ}-(20^{\circ} + 130^{\circ})\)
\(\alpha = 30^{\circ}\)

Step2: Use the Law of Sines to find \(a\)

The Law of Sines states \(\frac{a}{\sin A}=\frac{b}{\sin B}\). Here \(A = 20^{\circ}\), \(B=130^{\circ}\), \(b = 14\) in. So \(\frac{a}{\sin20^{\circ}}=\frac{14}{\sin130^{\circ}}\)
\(a=\frac{14\times\sin20^{\circ}}{\sin130^{\circ}}\)
\(\sin20^{\circ}\approx0.3420\), \(\sin130^{\circ}=\sin(180 - 50)^{\circ}=\sin50^{\circ}\approx0.7660\)
\(a=\frac{14\times0.3420}{0.7660}\approx6.3\) in

Step3: Use the Law of Sines to find \(c\)

\(\frac{c}{\sin\alpha}=\frac{14}{\sin130^{\circ}}\), since \(\alpha = 30^{\circ}\), \(\sin\alpha=\sin30^{\circ}=0.5\)
\(c=\frac{14\times\sin30^{\circ}}{\sin130^{\circ}}=\frac{14\times0.5}{0.7660}\approx9.1\) in

Answer:

\(\alpha = 30^{\circ},a = 6.3\) in., \(c = 9.1\) in. (First option)