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

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(Solution)
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==Solution==
 
==Solution==
{{solution}}
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Calculating for angles <math>\angle DBC</math> and <math>\angle DCB</math>, we get
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<math>\angle DBC</math> = <math>90 - \frac{B}{2}</math> and
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<math>\angle DCB</math> = <math>90 - \frac{C}{2}</math>.
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In triangle BCD, we have
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<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==
 
==See Also==

Revision as of 23:32, 28 August 2018

Problem

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:

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

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

Solution

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 $\fbox{C}$.

See Also

1962 AHSC (ProblemsAnswer KeyResources)
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

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