# Difference between revisions of "1995 AJHSME Problems/Problem 13"

Mrdavid445 (talk | contribs) (Created page with "==Problem== In the figure, <math>\angle A</math>, <math>\angle B</math>, and <math>\angle C</math> are right angles. If <math>\angle AEB = 40^\circ </math> and <math>\angle BED...") |
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<math>\text{(A)}\ 75^\circ \qquad \text{(B)}\ 80^\circ \qquad \text{(C)}\ 85^\circ \qquad \text{(D)}\ 90^\circ \qquad \text{(E)}\ 95^\circ</math> | <math>\text{(A)}\ 75^\circ \qquad \text{(B)}\ 80^\circ \qquad \text{(C)}\ 85^\circ \qquad \text{(D)}\ 90^\circ \qquad \text{(E)}\ 95^\circ</math> | ||

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+ | ==Solution== | ||

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+ | Because <math>\angle BED=\angle BDE</math>, <math>\angle B=90^\circ</math>, and <math>\triangle BED</math> is a triangle, we get: <cmath>\angle B + \angle BED +\angle BDE=180</cmath> <cmath>90 +\angle BED +\angle BED=180</cmath> <cmath>2\angle BED=90</cmath> <cmath>\angle BED=\angle BDE=45^\circ</cmath> | ||

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+ | So <math>\angle AED=\angle AEB +\angle BED=40 +45=85^\circ</math> Since ACDE is a quadrilateral, the sum of its angles is 360. Therefore: <cmath>\angle A +\angle C +\angle CDE +\angle AED=360</cmath> <cmath>90 +90 +\angle CDE +85=360</cmath> <cmath>\angle CDE=95^\circ \text{(E)}</cmath> |

## Revision as of 12:43, 5 July 2012

## Problem

In the figure, , , and are right angles. If and , then

## Solution

Because , , and is a triangle, we get:

So Since ACDE is a quadrilateral, the sum of its angles is 360. Therefore: