Question

A quadrilateral abcd is inscribed in a circle with centre o. If ∠boc = 92° and ∠adc = 112°, then ∠abo is equal

Answer 1

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Step-by-step explanation:

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Answer 2

$\angle ABO=24^{\circ}$

Step-by-step explanation:

$\angle BOC=92^{\circ}$

$\angle ADC=112^{\circ}$

The quadrilateral inscribed in a  circle therefore, it is cyclic quadrilateral because all vertices of quadrilateral lie on circle.

We know that

Sum of opposite angles of cyclic quadrilateral =180 degrees

$\angle ADC+\angle ABC=180^{\circ}$

Substitute the value then we get

$112+\angle ABC=180$

$\angle ABC=180-112=68^{\circ}$

In triangle BOC,

OC=OB

Radius of circle are equal

Let $\angle BCO=\angle CBO=x$

$\angle BOC=92^{\circ}$

Because when angle made by two equal sides are equal.

$\angle BOC+\angle CBO+\angle BCO=180$

Substitute the values then we get

$92+x+x=180$

$2x=180-92$

$2x=88$

$x=\frac{88}{2}=44^{\circ}$

$\angle ABO+\angle CBO=\angle ABC$

Substitute the values then we get

$\angle ABO+44=68$

$\angle ABO=68-44=24$

$\angle ABO=24^{\circ}$

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