+14915 In triangle ABC show above, cos B = 6/10. What is tan C?
Hello Guest!
\(c^2=a^2+b^2-2ab\cdot cos\ C\\ cos\ C=\dfrac{a^2+b^2-c^2}{2ab}\\ cos\ C=\dfrac{49+32-25}{56\cdot \sqrt{2}}=\dfrac{1}{\sqrt{2}}\)
\(C=\arccos\ \dfrac{1}{\sqrt{2}}\)
\(C=45°\)
\(tan\ C=tan 45°\)
\(tan\ C=1\)
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