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== | ||
+ | |||
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>. | ||
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+ | <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 23:13, 4 July 2013
Problem
If , and , then
Solution
The sum of the angles in a triangle is . We can find , so .
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 |
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