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1993 AHSME Problems/Problem 2: Difference between revisions

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== Solution ==
== Solution ==


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{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 15:37, 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 (Problems • Answer Key • Resources)
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

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