Question

In the given figure, PA and PB are tangent to a
circle with centre O and triangle ABC has been
inscribed in the circle such that AB = AC. If angle BAC =72
calculate (a) angle AOB (b) angle APB.
​

Answer 1

Answer:

Use the 2APB = AOB

Step-by-step explanation:

Because, the angle between 2 tangents are half of angle of center in circle.

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Answer 2

1.$m\angle AOB=108^{\circ}$

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

$x+72+x=180$

$2x=180-72=108$

$x=\frac{108}{2}=54$

$\angle ABC=\angle ACB=54^{\circ}$

Central angle is twice the inscribed angle

$m\angle AOB=2\times \angle ACB=2\times 54=108^{\circ}$

$m\angle AOB=108^{\circ}$

2.We know that radius is perpendicular to tangent

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

$\angle OAP=\angle OBA=90^{\circ}$

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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