1981 AHSME Problems/Problem 5

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

Draw the diagram using the information above. In triangle $DCB,$ note that $m\angle DCB=110^\circ$ and $m\angle CBD=30^\circ$ , so $m\angle CDB=40^\circ.$

Because $AB \parallel CD,$ we have $m\angle CDB= m\angle DBA = 40^\circ.$ Triangle $DAB$ is isosceles, so $m\angle ADB = 180 - 2(40) = 100^\circ.$

The answer is $\textbf{(C)}.$

-edited by coolmath34

1981 AHSME (Problems • Answer Key • Resources)
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1981 AHSME Problems/Problem 5

Problem 5

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