Question

question in \\( \triangle u v w, u=7 \mathrm{~cm}, w=3.7 \mathrm{~cm} \\) and \\( \angle w=19^{\circ} \\). find all possible values of \\( \angle u \\), to the nearest 10th of a degree. answer attempt 1 out of 2 additional angle no possible angles \\( \circ \\)

Explanation:

Step1: Use the Law of Sines

The Law of Sines states that \(\frac{u}{\sin U}=\frac{w}{\sin W}\). We know \(u = 7\), \(w=3.7\), and \(W = 19^{\circ}\). Substituting these values into the formula gives \(\frac{7}{\sin U}=\frac{3.7}{\sin19^{\circ}}\).

Step2: Solve for \(\sin U\)

Cross - multiply: \(3.7\sin U=7\sin19^{\circ}\). Then \(\sin U=\frac{7\sin19^{\circ}}{3.7}\). Calculate \(7\sin19^{\circ}\approx7\times0.3256 = 2.2792\). So \(\sin U=\frac{2.2792}{3.7}\approx0.616\).

Step3: Find the acute angle \(U\)

Using the inverse sine function, \(U_1=\sin^{- 1}(0.616)\approx38.0^{\circ}\).

Step4: Find the obtuse angle \(U\)

Since \(\sin\theta=\sin(180^{\circ}-\theta)\), the obtuse angle \(U_2 = 180^{\circ}-38.0^{\circ}=142.0^{\circ}\). Check if \(U_2\) is valid. The sum of angles in a triangle: \(U_2 + W=142^{\circ}+19^{\circ}=161^{\circ}<180^{\circ}\).

Answer:

\(38.0^{\circ}\) and \(142.0^{\circ}\)