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Difference between revisions of "1986 AHSME Problems/Problem 3"

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(Fixed the problem statement and added a solution with explanation)
 
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==Problem==
 
==Problem==
  
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<math>\triangle ABC</math> is a right angle at <math>C</math> and <math>\angle A = 20^\circ</math>. If <math>BD</math> (<math>D</math> in <math>\overline{AC}</math>) is the bisector of <math>\angle ABC</math>, then <math>\angle BDC =</math>
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<math>\triangle ABC</math> has a right angle at <math>C</math> and <math>\angle A = 20^\circ</math>. If <math>BD</math> (<math>D</math> in <math>\overline{AC}</math>) is the bisector of <math>\angle ABC</math>, then <math>\angle BDC =</math>
  
 
<math>\textbf{(A)}\ 40^\circ \qquad
 
<math>\textbf{(A)}\ 40^\circ \qquad
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==Solution==
 
==Solution==
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Since <math>\angle C = 90^{\circ}</math> and <math>\angle A = 20^{\circ}</math>, we have <math>\angle ABC = 70^{\circ}</math>. Thus <math>\angle DBC = 35^{\circ}</math>. It follows that <math>\angle BDC = 90^{\circ} - 35^{\circ} = 55^{\circ}</math>, which is <math>\boxed{D}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 17:03, 1 April 2018

Problem

$\triangle ABC$ has a right angle at $C$ and $\angle A = 20^\circ$. If $BD$ ($D$ in $\overline{AC}$) is the bisector of $\angle ABC$, then $\angle BDC =$

$\textbf{(A)}\ 40^\circ \qquad \textbf{(B)}\ 45^\circ \qquad \textbf{(C)}\ 50^\circ \qquad \textbf{(D)}\ 55^\circ\qquad \textbf{(E)}\ 60^\circ$

Solution

Since $\angle C = 90^{\circ}$ and $\angle A = 20^{\circ}$, we have $\angle ABC = 70^{\circ}$. Thus $\angle DBC = 35^{\circ}$. It follows that $\angle BDC = 90^{\circ} - 35^{\circ} = 55^{\circ}$, which is $\boxed{D}$.

See also

1986 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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All AHSME Problems and Solutions

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