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A triangle has sides with lengths of 10 miles, 12 miles, and 17 miles. Is it a right triangle? yes no

User Chicky
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you can solve a triangle if you have the 3 sides

using these formulas


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

So


\begin{gathered} \cos A=(10^2+12^2-17^2)/(2(10)(12)) \\ \cos A=(-45)/(240) \\ A=\cos ^(-1)((-45)/(240)) \\ A=100.8 \end{gathered}
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the sum of all the angles must equal 180


\begin{gathered} A+B+C=180 \\ 100.8+35.29+C=180 \\ C=180-35.29-100.8 \\ C=43.41 \end{gathered}

a rigth triangle have an angle of 90°, so this triangle isn't rigth

A triangle has sides with lengths of 10 miles, 12 miles, and 17 miles. Is it a right-example-1
User Discolor
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