Question

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Answer 1

Answer:

option b is correct

Step-by-step explanation:

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Answer 2

Answer:

ANSWER

In△PAQ

Sumoftwooppositeinteriorangles=Exteriorangle

α=θ+∠θ−(i)

In△PAR

α+θ+∠R=180

∘

α=180

∘

−θ−∠R−(ii)

In△PMQ

∠MPQ=180−∠Q−90

∘

=90−∠Q

∠APM=∠APQ−∠MPQ

=θ−(90−∠Q)

=θ+∠Q−90

∘

−(iii)

∵(i)=(ii)

=>θ+∠Q=180−θ−∠R

=>2θ=180−∠Q−∠R

=>θ=90−

2

1

(∠Q+∠R)

puttingthisvalueofθin(iii)

∠APM=∠Q+90−

2

1

(∠Q+∠R)−90

=

2

∠Q

−

2

∠R

=

2

1

(∠Q−∠R)

∴a=2