Difference between revisions of "1962 AHSME Problems/Problem 7"

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==Solution==
 
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==See Also==
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{{AHSME 40p box|year=1962|before=Problem 6|num-a=8}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Revision as of 21:14, 3 October 2014

Problem

Let the bisectors of the exterior angles at $B$ and $C$ of triangle $ABC$ meet at D$.$ Then, if all measurements are in degrees, angle $BDC$ equals:

$\textbf{(A)}\ \frac{1}{2}(90-A)\qquad\textbf{(B)}\ 90-A\qquad\textbf{(C)}\ \frac{1}{2}(180-A)\qquad$

$\textbf{(D)}\ 180-A\qquad\textbf{(E)}\ 180-2A$

Solution

See Also

1962 AHSC (ProblemsAnswer KeyResources)
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 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
All AHSME Problems and Solutions

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