Difference between revisions of "1993 AHSME Problems/Problem 2"

(Problem)
m
 
(One intermediate revision by one other user not shown)
Line 18: Line 18:
  
 
== Solution ==
 
== Solution ==
<math>\fbox{D}</math>
+
 
 +
We first consider <math>\angle CBA</math>. Because <math>\angle A = 55</math> and <math>\angle C = 75</math>, <math>\angle B = 180 - 55 - 75 = 50</math>. Then, because <math>\triangle BED</math> is isosceles, we have the equation <math>2 \angle BED + 50 = 180</math>. Solving this equation gives us <math>\angle BED = 65 \rightarrow \fbox{\textbf{(D)}65}</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 16:37, 14 July 2021

Problem

[asy] draw((-5,0)--(5,0)--(2,14)--cycle,black+linewidth(.75)); draw((-2.25,5.5)--(4,14/3),black+linewidth(.75)); MP("A",(-5,0),S);MP("C",(5,0),S);MP("B",(2,14),N);MP("E",(4,14/3),E);MP("D",(-2.25,5.5),W); MP("55^\circ",(-4.5,0),NE);MP("75^\circ",(5,0),NW);  [/asy]

In $\triangle ABC$, $\angle A=55^\circ$, $\angle C=75^\circ, D$ is on side $\overline{AB}$ and $E$ is on side $\overline{BC}$. If $DB=BE$, then $\angle{BED} =$


$\text{(A) } 50^\circ\quad \text{(B) } 55^\circ\quad \text{(C) } 60^\circ\quad \text{(D) } 65^\circ\quad \text{(E) } 70^\circ$

Solution

We first consider $\angle CBA$. Because $\angle A = 55$ and $\angle C = 75$, $\angle B = 180 - 55 - 75 = 50$. Then, because $\triangle BED$ is isosceles, we have the equation $2 \angle BED + 50 = 180$. Solving this equation gives us $\angle BED = 65 \rightarrow \fbox{\textbf{(D)}65}$

See also

1993 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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

Invalid username
Login to AoPS