Question

solve for the angles of the triangle described below. express all angles in degrees and round to the nearest hundredth.

( a = 9, b = 4, c = 6 )

Explanation:

Step1: Use the Law of Cosines to find angle \(A\)

The Law of Cosines formula is \(a^{2}=b^{2}+c^{2}-2bc\cos A\).
Substitute \(a = 9\), \(b = 4\), \(c = 6\) into the formula:
\(9^{2}=4^{2}+6^{2}-2\times4\times6\times\cos A\)
\(81 = 16 + 36-48\cos A\)
\(81=52 - 48\cos A\)
\(48\cos A=52 - 81=-29\)
\(\cos A=\frac{-29}{48}\approx - 0.6042\)
\(A=\cos^{-1}(-0.6042)\approx127.13^{\circ}\)

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

The Law of Cosines formula is \(b^{2}=a^{2}+c^{2}-2ac\cos B\)
Substitute \(a = 9\), \(b = 4\), \(c = 6\) into the formula:
\(4^{2}=9^{2}+6^{2}-2\times9\times6\times\cos B\)
\(16=81 + 36-108\cos B\)
\(16 = 117-108\cos B\)
\(108\cos B=117 - 16 = 101\)
\(\cos B=\frac{101}{108}\approx0.9352\)
\(B=\cos^{-1}(0.9352)\approx20.70^{\circ}\)

Step3: Use the angle - sum property of a triangle (\(A + B+C=180^{\circ}\)) to find angle \(C\)

\(C=180^{\circ}-A - B\)
\(C=180^{\circ}-127.13^{\circ}-20.70^{\circ}\)
\(C = 32.17^{\circ}\)

Answer:

\(A\approx127.13^{\circ}\), \(B\approx20.70^{\circ}\), \(C\approx32.17^{\circ}\)