Difference between revisions of "1994 AJHSME Problems/Problem 7"
Mrdavid445 (talk | contribs) (Created page with "If <math>\angle A = 60^\circ </math>, <math>\angle E = 40^\circ </math> and <math>\angle C = 30^\circ </math>, then <math>\angle BDC = </math> <asy> pair A,B,C,D,EE; A = origin;...") |
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| + | ==Problem== | ||
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If <math>\angle A = 60^\circ </math>, <math>\angle E = 40^\circ </math> and <math>\angle C = 30^\circ </math>, then <math>\angle BDC = </math> | If <math>\angle A = 60^\circ </math>, <math>\angle E = 40^\circ </math> and <math>\angle C = 30^\circ </math>, then <math>\angle BDC = </math> | ||
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<math>\text{(A)}\ 40^\circ \qquad \text{(B)}\ 50^\circ \qquad \text{(C)}\ 60^\circ \qquad \text{(D)}\ 70^\circ \qquad \text{(E)}\ 80^\circ</math> | <math>\text{(A)}\ 40^\circ \qquad \text{(B)}\ 50^\circ \qquad \text{(C)}\ 60^\circ \qquad \text{(D)}\ 70^\circ \qquad \text{(E)}\ 80^\circ</math> | ||
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| + | ==Solution== | ||
| + | We can find angle ABE is <math>80</math> degrees, so CBD is <math>180-80=100</math>. | ||
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| + | <math>180-100-30=</math> <math>\boxed{50}</math> | ||
Revision as of 13:35, 15 August 2011
Problem
If 
, 
and 
, then 
![[asy] pair A,B,C,D,EE; A = origin; B = (2,0); C = (5,0); EE = (1.5,3); D = (1.75,1.5); draw(A--C--D); draw(B--EE--A); dot(A); dot(B); dot(C); dot(D); dot(EE); label("$A$",A,SW); label("$B$",B,S); label("$C$",C,SE); label("$D$",D,NE); label("$E$",EE,N); [/asy]](http://latex.artofproblemsolving.com/f/0/d/f0dc899c415371f7deef8daea361750c75ff942f.png)
Solution
We can find angle ABE is 
degrees, so CBD is 
.
