Question

If 8R^2 = a^2 + b^2+c^2 ,prove that the triangle is right angle triangle. Here a,b and c are the lengths of side of triangle and R is circum radius.

Right triangles and trigonometry
ANSWERED
asked 2021-01-27
If \(\displaystyle{8}{R}^{{2}}={a}^{{2}}+{b}^{{2}}+{c}^{{2}}\) ,prove that the triangle is right angle triangle. Here a,b and c are the lengths of side of triangle and R is circum radius.

Answers (1)

2021-01-28
\(\displaystyle{a}={2}{r}{\sin{{A}}}\)
\(\displaystyle{b}={2}{r}{\sin{{B}}}\)
\(\displaystyle{c}={2}{r}{\sin{{C}}}\)
\(\displaystyle{a}^{{2}}+{b}^{{2}}+{c}^{{2}}={4}{R}^{{2}}{\left({{\sin}^{{2}}{A}}+{{\sin}^{{2}}{B}}+{{\sin}^{{2}}{C}}\right)}={8}{R}^{{2}}\)
\(\displaystyle{{\sin}^{{2}}{A}}+{{\sin}^{{2}}{B}}+{{\sin}^{{2}}{C}}={2}\)
\(\displaystyle{1}-{\cos{{2}}}{A}+{1}-{\cos{{2}}}{B}+{1}-{\cos{{2}}}{C}={4}\)
\(\displaystyle{\cos{{2}}}{A}+{\cos{{2}}}{B}+{\cos{{2}}}{C}+{1}={0}\)
\(\displaystyle{\cos{{A}}}=-{\cos{{\left({B}+{C}\right)}}}\)
\(\displaystyle{1}+{\cos{{2}}}{A}={2}{\cos{{\left({B}+{C}\right)}}}{\cos{{\left({B}+{C}\right)}}}\)
\(\displaystyle{2}{\cos{{\left({B}+{C}\right)}}}{\cos{{\left({B}+{C}\right)}}}+{\cos{{2}}}{B}+{\cos{{2}}}{C}={2}{\cos{{\left({B}+{C}\right)}}}{\cos{{\left({B}+{C}\right)}}}+{2}{\cos{{\left({B}+{C}\right)}}}{\cos{{\left({B}-{C}\right)}}}\)
\(\displaystyle{2}{\cos{{\left({B}+{C}\right)}}}{\left({\cos{{\left({B}+{C}\right)}}}+{\cos{{\left({B}-{C}\right)}}}\right)}={0}\)
\(\displaystyle{4}{\cos{{A}}}{\cos{{B}}}{\cos{{C}}}={0}\)
Some of the factors must be zero so one angle is right
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