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

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)
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

@@ Line 7: / Line 7: @@
 ==Solution==
-"Unsolved"
+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==
+{{AHSME 40p box|year=1962|before=Problem 6|num-a=8}}
+[[Category:Introductory Algebra Problems]]
+{{MAA Notice}}