Question

find the missing parts of the triangle. (find angles to the nearest hundredth of a degree.)
ab = 321 yd
ac = 186 yd
bc = 162 yd
a. a=21.12°, b=24.43°, c=134.45°
b. a=24.43°, b=21.12°, c=134.45°
c. a=21.12°, b=12.22°, c=146.66°
d. no triangle satisfies the given conditions.

Explanation:

Step1: Use the Law of Cosines

The Law of Cosines formula is \(c^{2}=a^{2}+b^{2}-2ab\cos C\). Let \(a = 162\), \(b = 186\), \(c = 321\).

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\(C=\cos^{-1}(-0.699)\approx134.45^{\circ}\)

Step2: Use the Law of Sines

The Law of Sines formula is \(\frac{\sin A}{a}=\frac{\sin C}{c}\). We know \(a = 162\), \(c = 321\), \(C\approx134.45^{\circ}\)

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\(A=\sin^{-1}(0.353)\approx21.12^{\circ}\)

Step3: Find angle B

Since \(A + B + C=180^{\circ}\), then \(B = 180^{\circ}-A - C\)
\(B=180^{\circ}-21.12^{\circ}-134.45^{\circ}=24.43^{\circ}\)

Answer:

A. \(A = 21.12^{\circ}\), \(B = 24.43^{\circ}\), \(C = 134.45^{\circ}\)