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I am attempting to solve a trigonometry problem that gives me the following information :

In $\triangle ABC$, if $a = 4$, $b = 5$, $c = 6$, compute $\tan C$.

I assumed that we draw a triangle with side lengths respectively as above (opposite to their angles) and got $6/5$ since tangent is the ratio of the opposite side over the adjacent side. However, this is not the answer. How do we approach this problem?

Thank you.

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2 Answers 2

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By the law of cosines $$\cos\measuredangle C=\frac{4^2+5^2-6^2}{2\cdot4\cdot5}=\frac{1}{8}$$ and since $\angle C$ is an acute angle, we obtain: $$1+\tan^2\measuredangle C=\frac{1}{\cos^2\measuredangle C}$$ or $$\tan\measuredangle C=\sqrt{63}.$$

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Hint:

Use $2ab\cos C=a^2+b^2-c^2$

Observe that $\sin C>0$ as $0<C<\pi$

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