1988 AJHSME Problems/Problem 5: Difference between revisions

Revision as of 22:55, 4 July 2013

Problem

If $\angle \text{CBD}$ is a right angle, then this protractor indicates that the measure of $\angle \text{ABC}$ is approximately

$[asy] unitsize(36); pair A,B,C,D; A=3*dir(160); B=origin; C=3*dir(110); D=3*dir(20); draw((1.5,0)..(0,1.5)..(-1.5,0)); draw((2.5,0)..(0,2.5)..(-2.5,0)--cycle); draw(A--B); draw(C--B); draw(D--B); label("O",(-2.5,0),W); label("A",A,W); label("B",B,S); label("C",C,W); label("D",D,E); label("0",(-1.8,0),W); label("20",(-1.7,.5),NW); label("160",(1.6,.5),NE); label("180",(1.7,0),E); [/asy]$

$\text{(A)}\ 20^\circ \qquad \text{(B)}\ 40^\circ \qquad \text{(C)}\ 50^\circ \qquad \text{(D)}\ 70^\circ \qquad \text{(E)}\ 120^\circ$

Solution

We have that $20^{\circ}+\angle ABD +\angle CBD=160^{\circ}$ , or $\angle ABD +\angle CBD=140^{\circ}$ . Since $\angle CBD$ is a right angle, we have $\angle ABD=140^{\circ}-90^{\circ}=50^{\circ}\Rightarrow \mathrm{(C)}$ .

1988 AJHSME (Problems • Answer Key • Resources)
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All AJHSME/AMC 8 Problems and Solutions

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 {{AJHSME box|year=1988|num-b=4|num-a=6}}
 [[Category:Introductory Geometry Problems]]
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1988 AJHSME Problems/Problem 5: Difference between revisions

Revision as of 22:55, 4 July 2013

Problem

Solution

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