Question

A triangleABC is scribed in a circle with centreO.if AB=AC,find angle ABC and angleABC​

Answer 1

Answer:

$1.m\angle AOB=108^{\circ} \\ </p><p>2.\angle APB=72^{\circ}$

Step-by-step explanation:

AB=AC

$\angle BAC=72^{\circ}$

Angle made by two equal sides are equal.

$\angle ABC=\angle ACB$

$Let \angle ABC=\angle ACB=x$

1.In triangle ABC

$\angle BAC+\angle ABC+\angle ACB=180$

By using triangle angle sum property

Substitute the values then we get

$x+72+x=180 \\ </p><p>2x=180-72=108 \\ </p><p>x=\frac{108}{2}=54 \\ </p><p>\angle ABC=\angle ACB=54^{\circ}$

Central angle is twice the inscribed angle

$m\angle AOB=2\times \angle ACB=2\times 54=108^{\circ}</p><p></p><p>m\angle AOB=108^{\circ}$

2.We know that radius is perpendicular to tangent

OA is perpendicular to PA and OB is perpendicular to PB.

$\angle OAP=\angle OBA=90^{\circ}</p><p>In quadrilateral OAPB$

In quadrilateral OAPB

$\angle OAP+\angle OBP+\angle APB+\angle AOB=360$

Substitute the values then we get

$90+90+108+\angle APB=360$

$\angle APB=360-288=72^{\circ}$

$\angle APB=72^{\circ}$

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