Difference between revisions of "1962 AHSME Problems/Problem 7"
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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== | ||
| − | + | 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}} | ||
Latest revision as of 00:34, 29 August 2018
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
Calculating for angles 
and 
, we get
= 
and
= 
.
In triangle 
, we have
= 
= 
= 
, so the answer is 
.
See Also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
