isosceles trapezoid ABCD,
AB = CD,
angle BDA = 49 degrees,
angle BDC = 13 degrees.
Find the degree measure of the angle ABD -?
1. Consider angle D. Angle D = angle CDB + angle BDA = 13 + 49 = 62 degrees.
2. Consider an isosceles trapezoid ABCD. Its angles at the base are equal to each other, that is, the angle A = angle D = 62 degrees.
3. Consider the triangle ABD. We know that the sum of the degree measures of the angles of a triangle is 180 degrees, that is
angle BAD + angle ADB + angle ABD = 180;
62 + 49 + angle ABD = 180;
111 + angle ABD = 180;
angle ABD = 180 – 111;
angle ABD = 69 degrees.
Answer: 69 degrees.
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