61. In a circle, with centrel O, a cyclic quadrilateal
ABCD is drawn with AB 3 as a diameter of the
circle and CD equal to radius of the circle. If
and BC produced meet at point P; show
that angle APB =
60°
please solve 61 number
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AB is a diameter and CD is equal to radius
so angle BAD = angle ABC= 60°
Now,
in triangle APB, we have
angle (A+B+C) =180°
60°+ B+60°=180°
B=180°-120°
B=60°
Hence, angle APB=60°
so angle BAD = angle ABC= 60°
Now,
in triangle APB, we have
angle (A+B+C) =180°
60°+ B+60°=180°
B=180°-120°
B=60°
Hence, angle APB=60°
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