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== | ||
| + | The sum of the angles in a triangle is <math>180^\circ</math>. We can find <math>\angle ABE = 80^\circ</math>, so <math>\angle CBD = 180-80=100^\circ</math>. | ||
| + | |||
| + | <cmath>\angle BDC = 180-100-30=\boxed{\text{(B)}\ 50^\circ}</cmath> | ||
| + | |||
| + | ==See Also== | ||
| + | {{AJHSME box|year=1994|num-b=6|num-a=8}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 00:13, 5 July 2013
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
The sum of the angles in a triangle is 
. We can find 
, so 
.
![\[\angle BDC = 180-100-30=\boxed{\text{(B)}\ 50^\circ}\]](http://latex.artofproblemsolving.com/b/4/4/b44902782037a74c37567d2dd12d73df313115d5.png)
See Also
| 1994 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
