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Difference between revisions of "2005 CEMC Gauss (Grade 7) Problems/Problem 8"

(Created page with "== Problem == In the diagram, what is the measure of <math>\angle ACB</math> in degrees? <asy> size(300); draw((-60,0)--(0,0)); draw((0,0)--(64.3,76.6)--(166,0)--cycle); label("...")
 
(→‎See Also)
 
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label("$C$",(166,0),S);
 
label("$C$",(166,0),S);
 
</asy>
 
</asy>
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<math>\text{(A)}\ 57^\circ \qquad \text{(B)}\ 37^\circ \qquad \text{(C)}\ 47^\circ \qquad \text{(D)}\ 60^\circ \qquad \text{(E)}\ 17^\circ</math>
  
 
== Solution ==
 
== Solution ==
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==See Also==
 
==See Also==
  
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[[2005 CEMC Gauss (Grade 7)]]
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{{CEMC box|year=2005|competition=Gauss (Grade 7)|num-b=7|num-a=9}}

Latest revision as of 00:46, 23 October 2014

Problem

In the diagram, what is the measure of $\angle ACB$ in degrees? [asy] size(300); draw((-60,0)--(0,0)); draw((0,0)--(64.3,76.6)--(166,0)--cycle); label("$A$",(64.3,76.6),N); label("$93^\circ$",(64.3,73),S); label("$130^\circ$",(0,0),NW); label("$B$",(0,0),S); label("$D$",(-60,0),S); label("$C$",(166,0),S); [/asy]

$\text{(A)}\ 57^\circ \qquad \text{(B)}\ 37^\circ \qquad \text{(C)}\ 47^\circ \qquad \text{(D)}\ 60^\circ \qquad \text{(E)}\ 17^\circ$

Solution

Since $\angle ABC + \angle ABD = 180^\circ$ (in other words, $\angle ABC$ and $\angle ABD$ are supplementary) and $\angle ABD = 130^\circ$, then $\angle ABC = 50^\circ$. Since the sum of the angles in triangle $ABC$ is $180^\circ$ and we know two angles $93^\circ$ and $50^\circ$ which add to $143^\circ$, then $\angle ACB = 180^\circ - 143^\circ = 37^\circ$. The answer is $B$.

See Also

2005 CEMC Gauss (Grade 7) (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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
CEMC Gauss (Grade 7)
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