Circles class 10....

1.	Circles class 10....
Answer» Assume angle OBD as x Make 2 constructions - OD and any point Q on the circle to join AQ and DQ Now, angle ODB is equal to OBD (isosceles triangle DOB) equal to x Angle AOD = ODB+OBD = 2x Angle AQD = 2x/2 = x (angle subtended on circle) AQDC is cyclic quadrilateral So angle ACD is 180-x (Result 1) Line segments OD, CD and OC are equal to radius, so ODC is equilateral triangle. Angle ODC is 60 deg. Angle CDP is THUS 180-60-x=120-x Now, angle CPD+angle CDP=angle ACD. Combining with Result 1, Angle CPD=180-x-120+x=60 degree Took less TIME to solve, but forever to write!

Answer»

Assume angle OBD as x

Make 2 constructions - OD and any point Q on the circle to join AQ and DQ

Now, angle ODB is equal to OBD (isosceles triangle DOB) equal to x

Angle AOD = ODB+OBD = 2x

Angle AQD = 2x/2 = x (angle subtended on circle)

AQDC is cyclic quadrilateral

So angle ACD is 180-x (Result 1)

Line segments OD, CD and OC are equal to radius, so ODC is equilateral triangle. Angle ODC is 60 deg. Angle CDP is THUS 180-60-x=120-x

Now, angle CPD+angle CDP=angle ACD. Combining with Result 1,

Angle CPD=180-x-120+x=60 degree

Took less TIME to solve, but forever to write!

Discussion