#Can anyone help me with the circles chapter class 9?
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~ABD = (~AOD)/2
~ABD = 90/2 = 45°
Reflex angle AOD + angle AOD = 360 [ because it's complete circle)
Reflex angle AOD = 360-90 = 270°
We can see quadrilateral ABDO
So, angle sum is 360°
~ABD + ~BDO + reflex ~DOA + ~OAB = 360
45° + ~BDO + 270° + ~OAB = 360°
So, ~BDO + ~OAB = 45°
So, by alternate interior angle therom
~ODB = y ```
Pervious equation will look like this
**x + y = 45°**
Now see triangle OBC which an isosceles triangle because OC and OB are radius
so, ~OBC = ~OCB = 30°
Hence, ~BOC + ~ OBC + ~ OCB = 180
~BOC = 180 - 60 = 120°
(~BOC)2 = ~BAC [Therom]
(120°)/2 = 2x
60° = 2x
**30° = x**
Remember, x + y = 45°
So, x = 30° ; y = 15°
-# " ~ " means angle, try to do it in simply way if you want short but complex you can tell me.
@fair otter
Thank you so much bro!!!!