In a trapezoid ABCK AB is perpendicular to AK, AC is perpendicular to CK, BC = 6 AK = 8 Find the angles of the trapezoid.
Let’s build the height of the CH.
Since the trapezoid is rectangular, and CH is its height, then the quadrilateral ABCH is a rectangle, then AH = BC = 6 cm.Then KH = AK – AH = 8 – 6 = 2 cm.
According to the condition AC is perpendicular to the SC, then the triangle ACK is rectangular, and CH is its height, drawn to the hypotenuse, and then CH ^ 2 = AH * KH.
CH ^ 2 = 6 * 2 = 12.
CH = 2 * √3 cm.
In a right-angled triangle SCH tgK = CH / KH = 2 * √3 / 2 = √3.
Angle AKC = arctg√3 = 60.
The sum of the angles of the trapezoid at the lateral side is 180, then the angle BCK = 180 – 60 = 120.
Answer: The angles of the trapezoid are 60, 90, 90, 120.
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