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pricebrittany41
18.09.2021 •
Mathematics
-3
2
The distance of AB rounded to the
nearest tenth = [?]
Use the distance formula: d = (x2 + x1)2 + (y2 + yı)2
Enter
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Ответ:
< CFE = 40°
Step-by-step explanation:
To better understand the solution, see attachment for the diagram.
Given:
BC parallel to DE
Measure of Arc BD = 58°
Measure of Arc DE = 142°
First step: Draw a diameter that passes through the centre of the circle and name it. In this case, the diameter is line ST.
The line ST divides the arc BD and arc DE into half.
That is:
Arc SC = 1/2(arc BC) =1/2(58)
Arc SC = 29°
Arc TE = 1/2(arc DE) =1/2(142)
Arc TE = 71°
Arc SC + Arc CE + Arc TE = 180° (Sum of angles in a semicircle
29° + Arc CE + 71° = 180°
Arc CE + 100° = 180°
Arc CE = 180-100
Arc CE = 80°
Inscribed angle = 1/2(intercepted angle)
<CFE = 1/2(Arc CE )
<CFE = 1/2(80)
< CFE = 40°