plz answer my question
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Angle AOC = 100° {given}
So, Angle APC = 100°/2 = 50° { 2:1 theorem }
We know that, ABCP is a cyclic quadrilateral
So, Angle APC + Angle ABC = 180°
=> Angle ABC = 130°
Now, Angle ABC + Angle CBD = 180° { linear pair }
Therefore, Angle CBD = 50°
Here, Angle APC = Angle CBD { since, exterior angle property of a cyclic quadrilateral }
So, we can directly write down Angle CBD = 50°
Hope this helps .....
So, Angle APC = 100°/2 = 50° { 2:1 theorem }
We know that, ABCP is a cyclic quadrilateral
So, Angle APC + Angle ABC = 180°
=> Angle ABC = 130°
Now, Angle ABC + Angle CBD = 180° { linear pair }
Therefore, Angle CBD = 50°
Here, Angle APC = Angle CBD { since, exterior angle property of a cyclic quadrilateral }
So, we can directly write down Angle CBD = 50°
Hope this helps .....
RanoMajhi:
can u plz solve in detail plzz
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