Question

consider a triangle abc like the one below. suppose that a = 36, b = 63, and c = 60. (the figure is not drawn to scale.) solve the triangle. carry your intermediate computations to at least four decimal places, and round your answers to the nearest tenth. if there is more than one solution, use the button labeled \or\. a = □ °, b = □ °, c = □ °

Explanation:

Step 1: Use the Law of Cosines

The Law of Cosines is \(a^{2}=b^{2}+c^{2}-2bc\cos A\), \(b^{2}=a^{2}+c^{2}-2ac\cos B\), \(c^{2}=a^{2}+b^{2}-2ab\cos C\). We can also use the Law of Sines \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\).
First, use the Law of Cosines to find \(\cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc}\).
Substitute \(a = 36\), \(b = 63\), \(c = 60\) into the formula:

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Then \(A=\cos^{-1}(0.83)\approx33.9^{\circ}\)

Step 2: Use the Law of Sines to find \(B\)

From the Law of Sines \(\frac{a}{\sin A}=\frac{b}{\sin B}\), so \(\sin B=\frac{b\sin A}{a}\)
Substitute \(a = 36\), \(b = 63\), \(A\approx33.9^{\circ}\)

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\(B=\sin^{-1}(0.975)\approx77.5^{\circ}\)

Step 3: Find \(C\)

Since \(A + B+C=180^{\circ}\), then \(C=180^{\circ}-A - B\)
Substitute \(A\approx33.9^{\circ}\), \(B\approx77.5^{\circ}\)
\(C=180-(33.9 + 77.5)=68.6^{\circ}\)

Answer:

\(A = 33.9^{\circ}\), \(B = 77.5^{\circ}\), \(C = 68.6^{\circ}\)