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question in \\( \triangle k l m, k = 94 \\) inches, \\( m = 79 \\) inches and \\( \angle m = 161 ^ { \circ } \\). find all possible values of \\( \angle k \\), to the nearest 10th of a degree. answer attempt 2 out of 2 additional angle no possible angles 22.8 submit answer watch video
Step1: Apply the Law of Sines
The Law of Sines states that \(\frac{k}{\sin K}=\frac{m}{\sin M}\). Substituting the given values \(k = 94\), \(m=79\), and \(M = 161^{\circ}\), we get \(\frac{94}{\sin K}=\frac{79}{\sin161^{\circ}}\).
Step2: Solve for \(\sin K\)
Cross - multiply: \(79\sin K=94\sin161^{\circ}\). Then \(\sin K=\frac{94\sin161^{\circ}}{79}\). Calculate \(\sin161^{\circ}\approx0.3256\), so \(\sin K=\frac{94\times0.3256}{79}\approx\frac{30.6064}{79}\approx0.3874\).
Step3: Find the reference angle
Since \(\sin K\approx0.3874\), the reference angle \(K_{ref}=\sin^{- 1}(0.3874)\approx22.8^{\circ}\).
Step4: Check for the second - quadrant angle (if applicable)
The sine function is positive in the first and second quadrants. But in a triangle, the sum of angles \(A + B + C=180^{\circ}\). If \(M = 161^{\circ}\), and we consider the second - quadrant angle \(K = 180^{\circ}-22.8^{\circ}=157.2^{\circ}\), then \(K + M=157.2^{\circ}+161^{\circ}=318.2^{\circ}>180^{\circ}\). So there is only one possible angle.
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\(22.8^{\circ}\)