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.
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°.