Question

solve the triangle below by finding all the missing sides and angles.

c = 17.8, a = 8.8°, b = 151.8°
c = 12.9, a = 12.1°, b = 148.5°
c = 4.47, a = 26.6°, b = 139.4°
c = 4.47a = 0.59°, b = 160°

Explanation:

Step1: Apply the Law of Sines

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). Here, let \(a = 8.25\), \(b = 12\), \(C=14^{\circ}\). First, \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\), so \(\frac{8.25}{\sin A}=\frac{12}{\sin B}=\frac{c}{\sin14^{\circ}}\). Also, \(A + B+14^{\circ}=180^{\circ}\), so \(B = 166^{\circ}-A\). Then \(\frac{8.25}{\sin A}=\frac{12}{\sin(166^{\circ}-A)}\). Using the sine - of - a - difference formula \(\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta\), \(\sin(166^{\circ}-A)=\sin166^{\circ}\cos A-\cos166^{\circ}\sin A=\sin(180 - 14)^{\circ}\cos A-\cos(180 - 14)^{\circ}\sin A=\sin14^{\circ}\cos A+\cos14^{\circ}\sin A\). So \(8.25(\sin14^{\circ}\cos A+\cos14^{\circ}\sin A)=12\sin A\). \(8.25\sin14^{\circ}\cos A=(12 - 8.25\cos14^{\circ})\sin A\). \(\tan A=\frac{8.25\sin14^{\circ}}{12 - 8.25\cos14^{\circ}}\approx\frac{8.25\times0.2419}{12 - 8.25\times0.9703}\approx\frac{1.9967}{12 - 8.005}\approx\frac{1.9967}{3.995}\approx0.5\). \(A\approx26.6^{\circ}\).

Step2: Find angle \(B\)

Since \(A + B + C=180^{\circ}\), \(B=180^{\circ}-(A + C)\). Substituting \(A = 26.6^{\circ}\) and \(C = 14^{\circ}\), \(B=180-(26.6 + 14)=139.4^{\circ}\).

Step3: Find side \(c\)

Using the Law of Sines \(\frac{c}{\sin C}=\frac{a}{\sin A}\). \(c=\frac{a\sin C}{\sin A}\). Substituting \(a = 8.25\), \(A = 26.6^{\circ}\), \(C = 14^{\circ}\), \(c=\frac{8.25\times\sin14^{\circ}}{\sin26.6^{\circ}}\). \(\sin14^{\circ}\approx0.2419\), \(\sin26.6^{\circ}\approx0.448\). \(c=\frac{8.25\times0.2419}{0.448}\approx4.47\).

Answer:

c = 4.47, \(A = 26.6^{\circ}\), \(B = 139.4^{\circ}\) (corresponds to the third option: \(c = 4.47,A = 26.6^{\circ},B = 139.4^{\circ}\))