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 to find angle \(A\)

The Law of Cosines formula is \(BC^{2}=AB^{2}+AC^{2}-2(AB)(AC)\cos A\).
Substitute \(AB = 321\), \(AC = 186\), \(BC = 162\) into the formula:
\(162^{2}=321^{2}+186^{2}-2\times321\times186\cos A\)
\(26244=103041 + 34596- 119172\cos A\)
\(26244=137637-119172\cos A\)
\(119172\cos A=137637 - 26244\)
\(119172\cos A=111393\)
\(\cos A=\frac{111393}{119172}\approx0.9347\)
\(A=\cos^{- 1}(0.9347)\approx20.99^{\circ}\approx21.12^{\circ}\)

Step2: Use the Law of Sines to find angle \(B\)

The Law of Sines formula is \(\frac{\sin B}{AC}=\frac{\sin A}{BC}\)
We know \(A\approx21.12^{\circ}\), \(AC = 186\), \(BC = 162\)
\(\sin B=\frac{AC\sin A}{BC}=\frac{186\times\sin(21.12^{\circ})}{162}\)
\(\sin B=\frac{186\times0.3607}{162}\)
\(\sin B=\frac{66.98}{162}\approx0.4135\)
\(B=\sin^{-1}(0.4135)\approx24.43^{\circ}\)

Step3: Find angle \(C\)

Since the sum of angles in a triangle is \(180^{\circ}\), \(C = 180^{\circ}-A - B\)
\(C=180^{\circ}-21.12^{\circ}-24.43^{\circ}=134.45^{\circ}\)

Answer:

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