Difference between revisions of "1963 AHSME Problems/Problem 34"

(Solution to Problem 34)
 
m (Solution)
 
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<cmath>\cos{x^\circ} < -\frac{1}{2}</cmath>
 
<cmath>\cos{x^\circ} < -\frac{1}{2}</cmath>
  
Note that <math>\cos{120^\circ} = -\frac{1}{2}</math>.  As <math>x</math> gets closer to <math>180^{\circ}</math>, <math>\cos{x}</math> decreases towards <math>-1</math>.  Thus, <math>x > 120</math>, so the answer is <math>\boxed{\textbf{(B)}}</math>
+
Note that <math>\cos{120^\circ} = -\frac{1}{2}</math>.  As <math>x</math> gets closer to <math>180^{\circ}</math>, <math>\cos{x}</math> decreases towards <math>-1</math>.  Thus, <math>x \leq 120</math>, and in order to maximize <math>x</math>, the answer has to be <math>\boxed{\textbf{(B)}}</math>
 
 
  
 
==See Also==
 
==See Also==

Latest revision as of 10:04, 15 July 2025

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, $x \leq 120$, and in order to maximize $x$, the answer has to be $\boxed{\textbf{(B)}}$

See Also

1963 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 33
Followed by
Problem 35
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
All AHSME Problems and Solutions

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