1957 AHSME Problems/Problem 44

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Problem

In $\triangle ABC, AC = CD$ and $\angle CAB - \angle ABC = 30^\circ$. Then $\angle BAD$ is:

[asy] defaultpen(linewidth(.8pt)); unitsize(2.5cm); pair A = origin; pair B = (2,0); pair C = (0.5,0.75); pair D = midpoint(C--B); draw(A--B--C--cycle); draw(A--D); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,NE);[/asy]

$\textbf{(A)}\ 30^\circ\qquad\textbf{(B)}\ 20^\circ\qquad\textbf{(C)}\ 22\frac{1}{2}^\circ\qquad\textbf{(D)}\ 10^\circ\qquad\textbf{(E)}\ 15^\circ$

Solution

Because $\triangle ACD$ is isosceles with $AC=CD$, $\measuredangle CAD=\measuredangle CDA=\theta$, where $\theta$ is some angle measure. Because $\angle CDA$ and $\angle ADB$ form a straight angle, $\measuredangle ADB = 180^{\circ}-\theta$. Thus, because the interior angles of a triangle add to $180^{\circ}$, $\measuredangle BAD + \measuredangle ABC = \theta$, so $\measuredangle ABC = \theta - \measuredangle BAD$. Notice that $\measuredangle CAB = \theta + \measuredangle BAD$. With all of this information and recalling that, from the problem, $\measuredangle CAB - \measuredangle ABC = 30^{\circ}$, we see that: CABABC=30θ+BAD(θBAD)=302BAD=30BAD=15 Thus, our answer is $\boxed{\textbf{(E) }15^{\circ}}$.

See Also

1957 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 43
Followed by
Problem 45
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