In the given figure , O is the centre of the circle, BD = OD and CD Is perpendicular to AB. Find angle CAB.
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We know that BD=OD.
But OD=OB= Radius of the circle.
Therefore
BD=OD=OB
BDO is equilateral triangle.
Angle DBO= 60°
Now let us take the intersecting point of CD and AB as X.
In triangle BDX,
BXD= 90°(BXD+BXC=180°, BXD+90°=180°, BXD=90°)
BXD+DBX+BDX=180°{Angle Sum Property}
90°+60°+BDX= 180°
BDX= 30°
We also know that,
BDC(BDX)= BAC (Angles lie on the same arc{BC} are equal in measure.
Therefore,
BAC=BDC(BDX)=30°
But OD=OB= Radius of the circle.
Therefore
BD=OD=OB
BDO is equilateral triangle.
Angle DBO= 60°
Now let us take the intersecting point of CD and AB as X.
In triangle BDX,
BXD= 90°(BXD+BXC=180°, BXD+90°=180°, BXD=90°)
BXD+DBX+BDX=180°{Angle Sum Property}
90°+60°+BDX= 180°
BDX= 30°
We also know that,
BDC(BDX)= BAC (Angles lie on the same arc{BC} are equal in measure.
Therefore,
BAC=BDC(BDX)=30°
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