1962 AHSME Problems/Problem 7
Problem
Let the bisectors of the exterior angles at 
and 
of triangle 
meet at D
Then, if all measurements are in degrees, angle 
equals:
$\textbf{(A)}\ \frac{1}{2}(90-A)\qquad\textbf{(B)}\ 90-A\qquad\textbf{(C)}\ \frac{1}{2}(180-A)\qquad\textbf{(D)}\ 180-A\qquad\textbf{(E)}\ \180-2A$ (Error compiling LaTeX. Unknown error_msg)
Solution
"Unsolved"
