# 1963 AHSME Problems/Problem 34

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## Problem

In $\triangle ABC$, side $a = \sqrt{3}$, side $b = \sqrt{3}$, and side $c > 3$. Let $x$ be the largest number such that the magnitude, in degrees, of the angle opposite side $c$ exceeds $x$. Then $x$ equals: $\textbf{(A)}\ 150^{\circ} \qquad \textbf{(B)}\ 120^{\circ}\qquad \textbf{(C)}\ 105^{\circ} \qquad \textbf{(D)}\ 90^{\circ} \qquad \textbf{(E)}\ 60^{\circ}$

## Solution

Using the Law of Cosines, $$\sqrt{3 + 3 - 2\cdot 3 \cdot \cos{x^\circ}}>3$$

Both sides are positive, so squaring both sides will not affect the inequality. $$6 - 6 \cos{x^\circ}>9$$ $$\cos{x^\circ} < -\frac{1}{2}$$

Note that $\cos{120^\circ} = -\frac{1}{2}$. As $x$ gets closer to $180^{\circ}$, $\cos{x}$ decreases towards $-1$. Thus, $x > 120$, so the answer is $\boxed{\textbf{(B)}}$

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