ADP and ABQ are tangents to the circle, centre O. C lies on the circumference of the circle. Giving reasons for all the statements you make, work out angles DCO, CBO, OCB and BCD, all in terms of x. Hence, work out the obtuse angle DOB and prove that y = 2x
Answers
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DCO=x (base angles in an isosceles are equal)
CBO=90-2x (radius meets tangent at 90 degrees)
OCB=90-2x (base angles in an isosceles are equal)
BCD=90-2x+x=90-x
DOB=180-2x (angles at the centre are double angles at the circumference)
Y=360-90-90-(180-2x)
Y=2x
Step-by-step explanation:
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Given : ADP and ABQ are tangents to the circle, centre O
To Find : work out angles DCO, CBO, OCB and BCD, all in terms of x
obtuse angle DOB and prove that y = 2x
Solution:
∠CBO = 180° - 90 - 2x = 90° - 2x
∠CBO = 90° - 2x
∠BCO = ∠CBO as OB = OC = Radius
∠BCO = 90° - 2x = ∠OCB
∠DCO = ∠CDO OC = OD = Radius
=> ∠DCO = x
∠BCD = ∠BCO + ∠DCO
= 90° - 2x + x
= 90° - x
∠BCD = 90° - x
∠DOB = 2∠BCD
=> ∠DOB = 2(90° - x)
=> ∠DOB = 180° - 2x
180° - 2x + 90° + 90° + y = 360° ( sum of angles of Quadrilateral)
=> -2x + y = 0
=> y = 2x
QED
Hence proved
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