1988 AJHSME Problems/Problem 5: Difference between revisions
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... |
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==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> | ||
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<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== | ||
[[ | {{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 
is a right angle, then this protractor indicates that the measure of 
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]](http://latex.artofproblemsolving.com/f/3/5/f35c58bf91cd15b3fe277437f0a010aeeef1f66d.png)
Solution
We have that 
, or 
. Since 
is a right angle, we have 
.
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 | ||
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