Question

determine the number of triangles abc possible with the given parts. a = 41° a = 8.5 b = 10.2 how many possible solutions does this triangle have?

Explanation:

Step1: Use the Law of Sines

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}\). Substituting the given values \(A = 41^{\circ}\), \(a = 8.5\), and \(b = 10.2\), we get \(\sin B=\frac{b\sin A}{a}\).
$$\sin B=\frac{10.2\times\sin41^{\circ}}{8.5}$$

Step2: Calculate \(\sin B\)

We know that \(\sin41^{\circ}\approx0.656\). Then \(\sin B=\frac{10.2\times0.656}{8.5}\).
First, calculate \(10.2\times0.656 = 6.6912\). Then \(\sin B=\frac{6.6912}{8.5}\approx0.7872\).

Step3: Analyze the number of solutions for \(B\)

Since \(\sin B\approx0.7872\), and \(0 < B<180^{\circ}\), there are two possible values for \(B\): \(B_1=\sin^{- 1}(0.7872)\approx51.9^{\circ}\) and \(B_2 = 180^{\circ}-51.9^{\circ}=128.1^{\circ}\).
For \(B_1 = 51.9^{\circ}\), \(C_1=180^{\circ}-(41^{\circ}+51.9^{\circ}) = 87.1^{\circ}\).
For \(B_2=128.1^{\circ}\), \(C_2=180^{\circ}-(41^{\circ}+128.1^{\circ})=10.9^{\circ}\).

Answer:

\(2\)