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In kite ABCD, mZBAE = 28° and mZBCE = 58°. Find mLABE.

QuestionExampleStep By StepIn Kite ABCD, MZBAE = 28 And MZBCE = 58. Find MLABE.

Answers

Answer 1

Answer:

[tex]m\angle ABE=52^\circ[/tex]

Step-by-step explanation:

Angles and Lines

The figure shows a kite ABCD where the following data is given:

[tex]m\BAE=28^\circ[/tex]

[tex]m\BCE=58^\circ[/tex]

We also know

[tex]BC\cong CD[/tex]

[tex]AB\cong AD[/tex]

Since triangle BCD is isosceles, it follows that:

[tex]m\angle CBE=m\angle CDE[/tex]

And triangles BCD and DCE are congruent.

Thus:

[tex]m\angle BCE=m\angle DCE=58^\circ[/tex]

And also:

[tex]m\angle BCE=58^\circ+58^\circ=116^\circ[/tex]

Since the sum of internal angles of BCD is 180°

[tex]m\angle CDE=m\angle CBE= ( 180 - 116 ) / 2=32^\circ[/tex]

The sum of angles of triangle CDE is 180°, thus

[tex]m\angle CED=180^\circ - 32^\circ - 58^\circ = 90^\circ[/tex]

Angles CED and BEA are vertical (opposite), thus:

[tex]m\angle BCE=m\angle CED= 90^\circ[/tex]

Finally, since the sum of the angles of triangle ABE is 180°

[tex]m\angle ABE= 180 - 90 - m\angle BAE[/tex]

[tex]m\angle ABE= 180^\circ - 90^\circ - 28^\circ = 52^\circ[/tex]

[tex]\mathbf{m\angle ABE=52^\circ}[/tex]


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Step-by-step explanation:

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Answers

9514 1404 393

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Step-by-step explanation:

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