1963 AHSME Problems/Problem 34

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 $\boxed{\textbf{(B)}}$

1963 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 33	Followed by Problem 35
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1963 AHSME Problems/Problem 34

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