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Sum of Opposite Angles of a Cyclic Quadrilateral

In the given ...

Question

In the given figure, AB is the diameter of a circle with centre O. $∠$ = $130^{o}$ . FInd : (i) $∠$ DAB (ii) $∠$ DBA

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Solution

(i) $\angle DAB = 180^o-\angle DCB = 180^o-130^o = 50^o \ (ABCD$ is a cyclic quadrilateral)

(ii) In $\triangle ADB$

$\angle DAB + \angle ADB + \angle DBA = 180^o$

$\Rightarrow \angle DBA = 180^o-50^o-90^o = 140^o$

$\to ∠ D B A = 180^{0} - 50^{0} - 90^{0} = 40^{0}$

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Similar questions

In the given figure, AB is the diameter of a circle with centre O. ∠BCD = 120⁰. Find ∠DBA.

In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and $∠$ DCB = $120^{o}$ . Calculate :

(i) $∠$ DAB, (ii) $∠$ DBA,

(iii) $∠$ DBC, (iii) $∠$ ADC.

Also, show that the $Δ$ AOD is an equilateral triangle.

∠ D A B = 25^{o}

∠ C A D .

A B

O

E D

A B

∠ E A B = 65^{o}

∠ E B A

∠ B E D

∠ B C D

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