Question

(3) In the figure, in A ABC
chord AB = chord AC
< BAC = 40°
then (1) m(arc BYC) = ?
(II) m (arc AXC) = ?​

Answer 1

Answer:

angle bac=40°

angle abc=angle acb = 70 ( given ab =ac,)

in a cyclic quadrilateral ABYC

angle bac + byc =180°

40°+byc=180°

byc=140°

in a cyclic quadrilateral axcb

angle abc + axc =180°

70°+axc = 180°

axc=180°-70°

axc=110°

Answer 2

1.$m(arc BYC)=80^{\circ}$

2.$m(arc AXC)=140$

Step-by-step explanation:

$\angle BAC=40^{\circ}$

$AB=AC$

$\angle ABC=\angle ACB$

Angle made by two equal sides are equal.

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

$\angle ABC+\angle ACB+\angle BAC=180^{\circ}$

By using triangle angle sum property

$x+x+40=80$

$2x=180-40=140$

$x=\frac{140}{2}=70^{\circ}$

We know that

Central angle is twice the inscribed angle

$\angle BOC=2\times \angle BAC=2\times 40=80^{\circ}$

Measure intercepted arc is equal to central angle

Therefore,$m(arc BYC)=80^{\circ}$

$\angle AOC=2\times \angle ABC=2\times 70=140^{\circ}$

$m(arc AXC)=\angle AOC=140$

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