Difference between revisions of "1994 AJHSME Problems/Problem 7"
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==Solution== | ==Solution== | ||
| − | We can find | + | 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}} | ||
Revision as of 23:21, 22 December 2012
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 | ||
