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In $\triangle ABC$, $$\angle A\;>\;\angle B\;>\;\angle C\;>\;\frac12 \angle A$$ Find the range for $\angle C$.

By

$$180^\circ=\angle A+\angle B+\angle C\;>\;\angle C+\angle C+\angle C\;=\;3\angle C$$

We can find the upper bound: $$\angle C\;<\;60^\circ$$

But how shall we find the lower bound?

Blue
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Cyh1368
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1 Answers1

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Well , $90^\circ=\frac{180^\circ}{2}=\frac{A}{2}+\frac{B}{2}+\frac{C}{2}\le C+C+\frac{C}{2}=\frac{5C}{2}$ $\Rightarrow C \ge \frac{90^\circ * 2}{5}=$ $\textbf{36}$ $^\circ$

Why is $\frac{A}{2}+\frac{B}{2}+\frac{C}{2}\le C+C+\frac{C}{2}$ ?

$\textbf{Answer :}$ $\frac{A}{2}\le C$ and $B\le A \Rightarrow \frac{B}{2} \le \frac{A}{2} \le C$

So , adding them up and adding $\frac{C}{2}$ on both sides , we get

$\frac{A}{2}+\frac{B}{2}+\frac{C}{2}\le C+C+\frac{C}{2}$

Vinny Dek
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