1981 AMC 12 Problem 5

Attempt Problem 5 of the 1981 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1981 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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=

80^\circ

90^\circ

100^\circ

110^\circ

120^\circ

Use the parallel bases to find the full interior angle at

B

Then use

AD=BD

in triangle

ABD

Since

AB\parallel CD,

consecutive interior angles give

\angle ABC=180^\circ-110^\circ=70^\circ.

Hence

\angle ABD=70^\circ-30^\circ=40^\circ.

Because

AD=BD,

triangle

ABD

has

\angle DAB=\angle ABD=40^\circ.

Therefore

\begin{aligned} \angle ADB &=180^\circ-40^\circ-40^\circ\\ &=100^\circ. \end{aligned}

Therefore, the correct answer is C.

Problem 5 in other years