Question

assume \\( \angle \alpha \\) is opposite side \\( a, \angle \beta \\) is opposite side \\( b \\), and \\( \angle \gamma \\) is opposite side \\( c \\).
solve triangle \\( a b c \\) if \\( \angle a=\alpha=38.9^{\circ}, b = 10.9 \mathrm{~km} \\), and \\( c = 14.4 \mathrm{~km} \\).
using the law of cosines,
\\( a \approx \\) km.
your answer should accurate to 2 decimal places.
using the law of cosines again,
\\( \cos \angle b=\cos \beta \approx \\)
your answer should accurate to 5 decimal places.
thus, \\( \beta \approx \\)
your answer should accurate to 2 decimal places.
\\( \cos \angle c=\cos \gamma \approx \\)
your answer should accurate to 5 decimal places.
thus, \\( \gamma \approx \\)
your answer should accurate to 2 decimal places.
check by seeing if the total of the three angles is \\( 180^{\circ} \\).

Explanation:

Step1: Find side \(a\) using the Law of Cosines

The Law of Cosines formula is \(a^{2}=b^{2}+c^{2}-2bc\cos\alpha\).
Substitute \(b = 10.9\), \(c = 14.4\), and \(\alpha=38.9^{\circ}\) (\(\cos(38.9^{\circ})\approx0.779\)):

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Step2: Find \(\cos\beta\) using the Law of Cosines

The formula \(\cos\beta=\frac{a^{2}+c^{2}-b^{2}}{2ac}\).
Substitute \(a\approx8.99\), \(b = 10.9\), \(c = 14.4\):

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Step3: Find \(\beta\)

Since \(\beta=\cos^{-1}(0.654)\), \(\beta\approx49.00^{\circ}\)

Step4: Find \(\cos\gamma\) using the Law of Cosines

The formula \(\cos\gamma=\frac{a^{2}+b^{2}-c^{2}}{2ab}\).
Substitute \(a\approx8.99\), \(b = 10.9\), \(c = 14.4\):

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Step5: Find \(\gamma\)

Since \(\gamma=\cos^{-1}(-0.039)\), \(\gamma\approx92.20^{\circ}\)

Answer:

\(a\approx8.99\) km, \(\cos\beta\approx0.654\), \(\beta\approx49.00^{\circ}\), \(\cos\gamma\approx - 0.039\), \(\gamma\approx92.20^{\circ}\)