In isosceles trapezoid ABCD, AB is parallel to DC. There are two diagnols that are in the trapezoid that intersect going from angle A to angle C and angle B to angle D. Angle BDC measures 25 degrees, and angle BCA measures 35 degrees. What is the measure of angle DBC?
The answer is somehow 95
The answer is somehow 95
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Let the intersection of the two diagonal lines be x. (I'm not incorporating the degrees sign.)
BCA is 25, and DCX is an isosceles triangle--> so, ACD is also 25
Thus,
DBC = 180 - 25 - (25 + 35)
DBC = 95
BCA is 25, and DCX is an isosceles triangle--> so, ACD is also 25
Thus,
DBC = 180 - 25 - (25 + 35)
DBC = 95
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Since ΔABC is congruent to ΔBCD, corresponding angles are equal.
Thus ∠DBC = ∠BCA = 35º
Thus ∠DBC = ∠BCA = 35º