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

Question Example Step by Step In kite ABCD, mZBAE = 28° and mZBCE = 58°. Find mLABE-example-1
User Njeri
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1 Answer

7 votes

Answer:


m\angle ABE=52^\circ

Explanation:

Angles and Lines

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


m\BAE=28^\circ


m\BCE=58^\circ

We also know


BC\cong CD


AB\cong AD

Since triangle BCD is isosceles, it follows that:


m\angle CBE=m\angle CDE

And triangles BCD and DCE are congruent.

Thus:


m\angle BCE=m\angle DCE=58^\circ

And also:


m\angle BCE=58^\circ+58^\circ=116^\circ

Since the sum of internal angles of BCD is 180°


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

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


m\angle CED=180^\circ - 32^\circ - 58^\circ = 90^\circ

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


m\angle BCE=m\angle CED= 90^\circ

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


m\angle ABE= 180 - 90 - m\angle BAE


m\angle ABE= 180^\circ - 90^\circ - 28^\circ = 52^\circ


\mathbf{m\angle ABE=52^\circ}

User Son
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