Here, ∠CDB = ∠BAC = 49° ∠ABC =∠ADC = 43° (Angle subtend by the same chord on the circle are equal) By angle – sum property of a triangle, ∠ACB = 180°- 49° -43°= 88°
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In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°.
Calculate : ∠DAB
Also show that the ΔAOD is an equilateral triangle .
In the figure, $O$ is the centre of the circle and the length of arc $AB$ is twice the length of arc $BC.$ If angle $AOB = 108^\circ$ find $: \angle ADB.$