Question

in the figure 'o' is the centre of the circle Angle OAB = 50° the measure of angle APC will be​

Answer 1

Answer:

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Step-by-step explanation:

so here I is the centre of adjoining circle

so bc is diameter of given circle

moreover ao and bo would be out radii

thus ao=bo (radii of same circle)

hence we can say that triangle oab is an isosceles triangle

hence by isosceles triangle theorem

we get

angle oab=angle oba=50 degrees

applying angle sum property in triangle aob

we get

angle aob=80 degrees

here angle aob is a central angle

so we know that

measure of central angle is same as that of arc intercepted by it ie subtended by it on circumference of circle

hence angle aob=m(arc ab)=80 degrees

thus we know that

Arc intercepted by diameter is always 180 and is called as a semi circle

hence measure of arc bc=180 degrees

now by arc addition property

we get

m(arc bc)+m(arc ab)+m(arc ac)=360

so m(arc ac)=360-(180+80)

=360-260

=100 degrees

so here angle apc intercepts arc ac on circumference of circle

thus by inscribed angle theorem

we get

angle apc=1/2×m(arc ac)

=1/2×100

=50

hence angle apc=50 degrees

thus option C is ryt