Difference between revisions of "1957 AHSME Problems/Problem 44"
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== Solution == | == Solution == | ||
− | <math>\boxed{\textbf{(E) }15^{\circ}}</math>. | + | 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} \\ | ||
+ | \theta + \measuredangle BAD - (\theta - \measuredangle BAD) &= 30^{\circ} \\ | ||
+ | 2\measuredangle BAD &= 30^{\circ} \\ | ||
+ | \measuredangle BAD &= 15^{\circ} | ||
+ | \end{align*} | ||
+ | Thus, our answer is <math>\boxed{\textbf{(E) }15^{\circ}}</math>. | ||
== See Also == | == See Also == | ||
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{{MAA Notice}} | {{MAA Notice}} | ||
[[Category:AHSME]][[Category:AHSME Problems]] | [[Category:AHSME]][[Category:AHSME Problems]] | ||
+ | [[Category:Introductory Geometry Problems]] |
Latest revision as of 11:30, 27 July 2024
Problem
In and
. Then
is:
Solution
Because is isosceles with
,
, where
is some angle measure. Because
and
form a straight angle,
. Thus, because the interior angles of a triangle add to
,
, so
. Notice that
. With all of this information and recalling that, from the problem,
, 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
.
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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