Question

6. PAQ is tangent to a circle with centre O at
point A. If angle OBA = 40, angleBOA = 105, then
find angleBAP​

Answer 1

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In ΔOAB

∵OA=OB (radii of a circle)

∴∠OBA=∠OAB (Opposite angle of the equal side of triangle)

∴∠OAB=32

∘

⇒∠x=32

∘

(∠OBA=32

∘

) ....(i)

Again, PAQ the tangent at point A of circle and OA is radius.

∴OA⊥PQ

⇒∠OAQ=90

∘

∴∠BAQ+∠OAB=90

∘

∴∠BAQ+32

∘

=90

∘

[∵∠OAB=32

∘

]

∠BAQ=90

∘

−32

∘

=58

∘

∠BAQ=∠ACB [ angle made in alternate segment]

58

∘

=∠y

∠x=32

∘

,∠y=58

∘

solution