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Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698

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BuyFindarrow_forward

Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698
Textbook Problem
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Given: m D C A = 130 ° m B A C = 2 x + y m B C E = 150 ° m D E C = 2 x y
Find: x and y

Chapter 2.CR, Problem 5CR, Given: mDCA=130mBAC=2x+ymBCE=150mDEC=2xy Find: x and y

To determine

To find:

The values of x and y.

Explanation

Given:

The given statement is,

mDCA=130°mBAC=2x+ymBCE=150°mDEC=2xy

And,

AB¯CD¯ and BC¯DE¯.

Figure (1)

Property:

(1) If two parallel lines are cut by a transversal, then a pair of alternate interior angles is congruent.

(2) The measure of an exterior angle of a triangle equals the sum of the measures of the two nonadjacent interior angles.

(3) Angles on a straight line add up to 180°.

Calculation:

The given statement is,

mDCA=130°mBAC=2x+ymBCE=150°mDEC=2xy

And,

AB¯CD¯ and BC¯DE¯.

mDCA+mBCE+mBCD=180°

Substitute 130° for mDCA and 150° for mBCE in the above equation.

130°+150°mBCD=180°mBCD=130°+150180°mBCD=100°

B and BCD are alternate interior angles.

BBCDmB=mBCD

Substitute 100° for mBCD in the above equation

mB=100°

BCD and D are alternate interior angles.

BCDDmBCD=mD

Substitute 100° for mBCD in the above equation

mD=100°

BCE is an exterior angle of triangle ΔABC

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