Asked by Ncm
In the figure below, not drawn to scale, ABC is an isosceles triangle with AC equal CB and CDE is an equilateral triangle. Given that BAC equal 70° and BEC is a straight line, find DCA.
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Answered by
GPT 3.5
To find the measure of angle DCA, we start by recognizing that angle C is equal to 180° - 70° = 110°. Since triangle ABC is isosceles, angles CBA and CAB are equal. Thus, angle CBA = angle CAB = (180° - 70°)/2 = 55°.
In triangle CBE, angle BCE is equal to 180° - 55° = 125°. Since triangle CDE is equilateral, all sides are equal which implies all angles are equal. Therefore, angle CED = angle CDE = angle DCE = 60°.
Now, in triangle CDE, the sum of the angles is 180°. Therefore, angle CED + angle CDE + angle DCE = 180°, which gives 60° + 60° + DCA = 180°. Solving for DCA, we find:
60° + 60° + DCA = 180°
120° + DCA = 180°
DCA = 180° - 120°
DCA = 60°
Therefore, DCA is equal to 60°.
In triangle CBE, angle BCE is equal to 180° - 55° = 125°. Since triangle CDE is equilateral, all sides are equal which implies all angles are equal. Therefore, angle CED = angle CDE = angle DCE = 60°.
Now, in triangle CDE, the sum of the angles is 180°. Therefore, angle CED + angle CDE + angle DCE = 180°, which gives 60° + 60° + DCA = 180°. Solving for DCA, we find:
60° + 60° + DCA = 180°
120° + DCA = 180°
DCA = 180° - 120°
DCA = 60°
Therefore, DCA is equal to 60°.
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