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Solve the triangle A = 2 B = 9 C =8

Solve the triangle A = 2 B = 9 C =8-example-1

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Answer:


\begin{gathered} A=\text{ 12}\degree \\ B=\text{ 114}\degree \\ C=54\degree \end{gathered}

Explanation:

To calculate the angles of the given triangle, we can use the law of cosines:


\begin{gathered} \cos (C)=(a^2+b^2-c^2)/(2ab) \\ \cos (A)=(b^2+c^2-a^2)/(2bc) \\ \cos (B)=(c^2+a^2-b^2)/(2ca) \end{gathered}

Then, given the sides a=2, b=9, and c=8.


\begin{gathered} \cos (A)=(9^2+8^2-2^2)/(2\cdot9\cdot8) \\ \cos (A)=(141)/(144) \\ A=\cos ^(-1)((141)/(144)) \\ A=11.7 \\ \text{ Rounding to the nearest degree:} \\ A=12º \end{gathered}

For B:


\begin{gathered} \cos (B)=(8^2+2^2-9^2)/(2\cdot8\cdot2) \\ \cos (B)=(13)/(32) \\ B=\cos ^(-1)((13)/(32)) \\ B=113.9\degree \\ \text{Rounding:} \\ B=114\degree \end{gathered}
\begin{gathered} \cos (C)=(2^2+9^2-8^2)/(2\cdot2\cdot9) \\ \cos (C)=(21)/(36) \\ C=\cos ^(-1)((21)/(36)) \\ C=54.3 \\ \text{Rounding:} \\ C=\text{ 54}\degree \end{gathered}

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