Question

what is the answer of this question​

Answer 1

Answer:

hey here is your answer

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both methods are applicable

Step-by-step explanation:

Method 1=

here angle abc=30 degrees

thus then angle abq=30 degrees (b-c-q)

so here angle bap is an exterior angle for triangle bqa here

thus by exterior angle theorem

we get

angle bap=angle bqa+angle abq

ie 80=angle bqa+30

so angle bqa=50 degrees

ie angle cqa=50 degree (b-c-q)

thus option A is correct

Method 2=

here angle abc intercepts arc ac on circumference of circle

so by inscribed angle theorem

we get

angle abc=1/2×m(arc ac)

ie m(arc ac)=2×30

so m(arc ac)=60 degrees

similarly angle here paq ie pa is a tangent and ab is a secant of adjoining circle

and angle bap is inscribed in between both of these

thus by tangent secant angle theorem

we get

angle bap=1/2×m(arc ab)

ie m(arc ab)=80×2

=160 degrees

so here in between angle bqp two arcs namely

m(arc ac) and m(arc ab) are intercepted ie are inscribed

so this then angle bqp=1/2×[m(arc ab)-m(arc ac)]

=1/2×(160-60)

=1/2×100

=50 degrees

hence my collinearity of p-a-q and b-c-q

we get

angle cqa ie angle aqc=50 degrees