In the given figure- O is the centre of the circle- If BD - DC and - DBC -30- find the measures of - BAC and - BOC-

BD = DC So, Δ BDC is an isosceles triangle BD = DC ∠ DBC = ∠ DCB = 30^∘ ∴∠ BDC = 120^∘ (angle seen property of Δ) ABDC is a cyclic quadrilateral So, ∠ BAC = 1 ...

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In the given ...

Question

In the given figure, O is the centre of the circle. If BD = DC and $∠ D B C = 30^{\circ}$ , find the measures of $∠ B A C$ and $∠ B O C$ .

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Solution

BD = DC
So, $Δ B D C$ is an isosceles triangle
BD = DC
$∠ D B C = ∠ D C B = 30^{\circ}$
$∴ ∠ B D C = 120^{\circ}$ (angle seen property of $Δ$ )
ABDC is a cyclic quadrilateral
So, $∠ B A C = 180^{\circ} - 120^{\circ} = 60^{\circ}$ (opposite angle is supplementary in cyclic quadrilateral)
$∠ B O C = 2 \times ∠ B A C = 2 \times 60^{\circ}$
$= 120^{\circ}$ (Angle subtend at centre is twice of angle subtend remaining part of circle)

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In the given figure, O is the centre of the circle. If BD = DC and ∠DBC=30∘, find the measures of ∠BAC and ∠BOC.

In the given figure, O is the centre of the circle. If BD = DC and $∠ D B C = 30^{\circ}$ , find the measures of $∠ B A C$ and $∠ B O C$ .