Find the corners of a rectangular trapezoid, the diagonal of which divides the obtuse angle of the trapezoid

| September 8, 2021 | education

Find the corners of a rectangular trapezoid, the diagonal of which divides the obtuse angle of the trapezoid in half and forms an angle of 20 ° with its smaller lateral side.

Let ABCD be a given rectangular trapezoid (AD and BC are bases), angle A = angle B = 90 ° (rectangular trapezoid). BAC angle = 20 °, BCA angle = DCA angle.

BCA angle = CAB angle (as internal opposing angles with parallel AD and BC and secant AC).

Angle BAD = CAB + DAC = 90 °.

Hence the angle CAB = 90 ° – 20 ° = 70 °.

Consequently, the angle of the BCA = 70 °. Since BCA = angle DCA = 70 °, the angle C of the trapezoid ABCD is 70 ° + 70 ° = 140 °.

The angles of the trapezium, adjacent to one side, add up to 180 °, which means that the angle D of the trapezoid ABCD is 180 ° – 140 ° = 40 °.

Answer: The angles of the trapezoid are 90 °, 90 °, 140 ° and 40 °.



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