Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1988 AJHSME Problems/Problem 5"

(New page: ==Problem== If <math>\angle \text{CBD}</math> is a right angle, then this protractor indicates that the measure of <math>\angle \text{ABC}</math> is approximately <asy> unitsize(36); pair...)
 
m (→‎Solution)
 
(2 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
−
If <math>\angle \text{CBD}</math> is a right angle, then this protractor indicates that the measure of <math>\angle \text{ABC}</math> is approximately
+
If <math>\angle \text{CBD}</math> is a [[right angle]], then this protractor indicates that the measure of <math>\angle \text{ABC}</math> is approximately
  
 
<asy>
 
<asy>
Line 22: Line 22:
 
<math>\text{(A)}\ 20^\circ \qquad \text{(B)}\ 40^\circ \qquad \text{(C)}\ 50^\circ \qquad \text{(D)}\ 70^\circ \qquad \text{(E)}\ 120^\circ</math>
 
<math>\text{(A)}\ 20^\circ \qquad \text{(B)}\ 40^\circ \qquad \text{(C)}\ 50^\circ \qquad \text{(D)}\ 70^\circ \qquad \text{(E)}\ 120^\circ</math>
 
==Solution==
 
==Solution==
−
We have that <math>20^{\circ}+\angle ABD +\angle CBD=160^{\circ}</math>, or <math>\angle ABD +\angle CBD=140^{\circ}</math>. Since <math>\angle CBD</math> is a right angle, we have <math>\angle ABD=140^{\circ}-90^{\circ}=50^{\circ}\Rightarrow \mathrm{(C)}</math>.
+
We have that <math>20^{\circ}+\angle ABD +\angle CBD=160^{\circ}</math>, or <math>\angle ABD +\angle CBD=140^{\circ}</math>. Since <math>\angle CBD</math> is a right angle, we have <math>\angle ABD=140^{\circ}-90^{\circ}=50^{\circ}\Rightarrow \boxed{\mathrm{(C)}}</math>.
  
 
==See Also==
 
==See Also==
  
−
[[1988 AJHSME Problems]]
+
{{AJHSME box|year=1988|num-b=4|num-a=6}}
 +
[[Category:Introductory Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 15:53, 17 December 2020

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 $\angle ABD=140^{\circ}-90^{\circ}=50^{\circ}\Rightarrow \boxed{\mathrm{(C)}}$.

See Also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1988_AJHSME_Problems/Problem_5&oldid=139872"