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Difference between revisions of "1981 AHSME Problems/Problem 5"

(Original problem included a diagram.)
 
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==Solution==
 
==Solution==
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Draw the diagram using the information above. In triangle <math>DCB,</math> note that <math>m\angle DCB=110^\circ </math> and <math> m\angle CBD=30^\circ </math>, so <math>m\angle CDB=40^\circ.</math>
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In <math>\triangle BCD</math>, <math>\angle CDB=180^\circ - \angle DCB - \angle CBD = 180^\circ - 30^\circ - 110^\circ = 40^\circ.</math>
  
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Because <math>AB \parallel CD,</math> we have <math>m\angle CDB= m\angle DBA = 40^\circ.</math> Triangle <math>DAB</math> is isosceles, so <math>m\angle ADB = 180 - 2(40) = 100^\circ.</math>
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Because <math>AB \parallel CD</math>, <math>\angle CDB= \angle DBA = 40^\circ</math>. Since <math>\triangle DAB</math> is isosceles, <math>\angle ADB = 180^\circ - 2 \cdot 40^\circ = 100^\circ.</math> <math>\fbox{(C)}</math>
  
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The answer is <math>\textbf{(C)}.</math>
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-edited by coolmath34, j314andrews
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-edited by coolmath34
 
  
 
== See Also ==
 
== See Also ==
 
{{AHSME box|year=1981|num-b=4|num-a=6}}
 
{{AHSME box|year=1981|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 12:14, 29 June 2025

Problem 5

In trapezoid $ABCD$, sides $AB$ and $CD$ are parallel, and diagonal $BD$ and side $AD$ have equal length. If $\angle DCB=110^\circ$ and $\angle CBD=30^\circ$, then $\angle ADB=$

[asy] import geometry; size(8cm); pair A=(0,0), D=(1.1918,1), B=(2.3835,0), C=(2.0195,1); draw(A--B--C--D--cycle, black); draw(B--D, black); draw(A--D, StickIntervalMarker(1,1)); draw(D--B, StickIntervalMarker(1,1)); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, NW); label("$110^\circ$", C, .6S+.5W); label("$?$", D, 1.5S); label("$30^\circ$", B, 7.5N+4.7W); [/asy]

$\textbf{(A)}\ 80^\circ\qquad\textbf{(B)}\ 90^\circ\qquad\textbf{(C)}\ 100^\circ\qquad\textbf{(D)}\ 110^\circ\qquad\textbf{(E)}\ 120^\circ$

Solution

In $\triangle BCD$, $\angle CDB=180^\circ - \angle DCB - \angle CBD = 180^\circ - 30^\circ - 110^\circ = 40^\circ.$

Because $AB \parallel CD$, $\angle CDB= \angle DBA = 40^\circ$. Since $\triangle DAB$ is isosceles, $\angle ADB = 180^\circ - 2 \cdot 40^\circ = 100^\circ.$ $\fbox{(C)}$

-edited by coolmath34, j314andrews

See Also

1981 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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
All AHSME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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