1962 AHSME Problems/Problem 7: Difference between revisions
| Line 2: | Line 2: | ||
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)} | <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> | ||
==Solution== | ==Solution== | ||
"Unsolved" | "Unsolved" | ||
Revision as of 21:29, 9 November 2013
Problem
Let the bisectors of the exterior angles at 
and 
of triangle 
meet at D
Then, if all measurements are in degrees, angle 
equals:
Solution
"Unsolved"
