1962 AHSME Problems/Problem 7: Difference between revisions

Latest revision as of 23:34, 28 August 2018

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:

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

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

Calculating for angles $\angle DBC$ and $\angle DCB$ , we get

$\angle DBC$ = $90 - \frac{B}{2}$ and $\angle DCB$ = $90 - \frac{C}{2}$ .

In triangle $BCD$ , we have

$\angle BDC$ = $180 - (90 - \frac{B}{2}) - (90 - \frac{C}{2})$ = $\frac{B+C}{2}$ = $\frac{1}{2}\cdot(180 - A)$ , so the answer is $\fbox{C}$ .

1962 AHSC (Problems • Answer Key • Resources)
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All AHSME Problems and Solutions

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 ==Solution==
+Calculating for angles <math>\angle DBC</math> and <math>\angle DCB</math>, we get
+<math>\angle DBC</math> = <math>90 - \frac{B}{2}</math> and
+<math>\angle DCB</math> = <math>90 - \frac{C}{2}</math>.
+In triangle <math>BCD</math>, we have
+<math>\angle BDC</math> = <math>180 - (90 - \frac{B}{2}) - (90 - \frac{C}{2})</math> = <math>\frac{B+C}{2}</math> = <math>\frac{1}{2}\cdot(180 - A)</math>, so the answer is <math>\fbox{C}</math>.
 ==See Also==