Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Created page with "==Problem== <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>)...")
 
(Fixed the problem statement and added a solution with explanation)
 
(One intermediate revision by one other user not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
  
−
<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>
+
<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
Line 10: Line 10:
  
 
==Solution==
 
==Solution==
−
 
+
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 ==
−
{{AHSME box|year=1986|before=num-b=2|num-a=4}}   
+
{{AHSME box|year=1986|num-b=2|num-a=4}}   
  
 
[[Category: Introductory Geometry Problems]]
 
[[Category: Introductory Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:03, 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
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
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1986_AHSME_Problems/Problem_3&oldid=93687"