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

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

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

Problem 5

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