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

(Problem)
(Problem)
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Let the bisectors of the exterior angles at <math>B</math> and <math>C</math> of triangle <math>ABC</math> meet at D<math>.</math> Then, if all measurements are in degrees, angle <math>BDC</math> equals:  
 
Let the bisectors of the exterior angles at <math>B</math> and <math>C</math> of triangle <math>ABC</math> meet at D<math>.</math> Then, if all measurements are in degrees, angle <math>BDC</math> equals:  
  
<math> \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</math>
+
<math> \textbf{(A)}\ \frac{1}{2}(90-A)\qquad\textbf{(B)}\ 90-A\qquad\textbf{(C)}\ \frac{1}{2}(180-A)\qquad</math>
 +
 
 +
<math>\textbf{(D)}\ 180-A\qquad\textbf{(E)}\ 180-2A</math>
  
 
==Solution==
 
==Solution==
 
"Unsolved"
 
"Unsolved"

Revision as of 21:58, 9 November 2013

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

"Unsolved"