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

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== See Also ==
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{{AHSME box|year=1981|num-b=4|num-a=6}}
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{{MAA Notice}}

Revision as of 13:31, 28 June 2025

Problem 5

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

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

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

See Also

1981 AHSME (ProblemsAnswer KeyResources)
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

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