Question

question
in the diagram below, \\( \overline { b e } \cong \overline { e c } \\), \\( \mathrm { m } \angle b e c = 52 ^ { \circ } \\) and \\( \mathrm { m } \angle a = 43 ^ { \circ } \\). find \\( \mathrm { m } \angle a e b \\).
step
angle
reason
1
\\( \mathrm { m } \angle b e c = 52 ^ { \circ } \\)
given
2
\\( \mathrm { m } \angle a = 43 ^ { \circ } \\)
given
try \\( \mathrm { m } \angle \square = \square ^ { \circ } \\)
select a reason

you may assume lines that appear straight are straight, but the figure is not otherwise drawn to scale.

Explanation:

Step1: Find the base angles of isosceles triangle \( \triangle BEC \)

Since \( \overline{BE}\cong\overline{EC} \), \( \triangle BEC \) is isosceles. Using the angle - sum property of a triangle (\( \angle BEC+\angle EBC+\angle ECB = 180^{\circ} \)) and \( \angle EBC=\angle ECB \) (base angles of an isosceles triangle). Let \( \angle EBC = \angle ECB=x \). Then \( 52^{\circ}+2x = 180^{\circ} \), so \( 2x=180^{\circ}-52^{\circ}=128^{\circ} \), and \( x = 64^{\circ} \). Also, \( \angle EBA = 180^{\circ}-\angle EBC \) (linear - pair of angles). So \( \angle EBA=180^{\circ}-64^{\circ}=116^{\circ} \).

Step2: Use the angle - sum property of \( \triangle AEB \)

In \( \triangle AEB \), using the angle - sum property (\( \angle A+\angle EBA+\angle AEB = 180^{\circ} \)). We know \( \angle A = 43^{\circ} \) and \( \angle EBA = 116^{\circ} \). Substitute into the formula: \( 43^{\circ}+116^{\circ}+\angle AEB=180^{\circ} \). Then \( \angle AEB=180^{\circ}-(43^{\circ}+116^{\circ}) \).

Step3: Calculate \( \angle AEB \)

\( \angle AEB=180^{\circ}-159^{\circ}=21^{\circ} \)

Answer:

\( m\angle AEB = 21^{\circ} \)