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Difference between revisions of "1957 AHSME Problems/Problem 44"

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== Solution ==
 
== Solution ==
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<math>\boxed{\textbf{(E) }15^{\circ}}</math>.
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Because <math>\triangle ACD</math> is [[Isosceles triangle|isosceles]] with <math>AC=CD</math>, <math>\measuredangle CAD=\measuredangle CDA=\theta</math>, where <math>\theta</math> is some angle measure. Because <math>\angle CDA</math> and <math>\angle ADB</math> form a [[straight angle]], <math>\measuredangle ADB = 180^{\circ}-\theta</math>. Thus, because the interior angles of a triangle add to <math>180^{\circ}</math>, <math>\measuredangle BAD + \measuredangle ABC = \theta</math>, so <math>\measuredangle ABC = \theta - \measuredangle BAD</math>. Notice that <math>\measuredangle CAB = \theta + \measuredangle BAD</math>. With all of this information and recalling that, from the problem, <math>\measuredangle CAB - \measuredangle ABC = 30^{\circ}</math>, we see that:
 +
\begin{align*}
 +
\measuredangle CAB - \measuredangle ABC &= 30^{\circ} \\
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\theta + \measuredangle BAD - (\theta - \measuredangle BAD) &= 30^{\circ} \\
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2\measuredangle BAD &= 30^{\circ} \\
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\measuredangle BAD &= 15^{\circ}
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\end{align*}
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Thus, our answer is <math>\boxed{\textbf{(E) }15^{\circ}}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 10:30, 27 July 2024

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: \begin{align*} \measuredangle CAB - \measuredangle ABC &= 30^{\circ} \\ \theta + \measuredangle BAD - (\theta - \measuredangle BAD) &= 30^{\circ} \\ 2\measuredangle BAD &= 30^{\circ} \\ \measuredangle BAD &= 15^{\circ} \end{align*} Thus, our answer is $\boxed{\textbf{(E) }15^{\circ}}$.

See Also

1957 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 43 		Followed by
Problem 45
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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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