Question

AB is a diameter
PQ is a chord
if angle APB = 40°
find x . sorry I didn't mention the name of the points in the previous question​

Answer 1

Answer:

Given, ∠AOB=110

∘

In △AOB

OA=OB (Radius of the circle)

Thus, ∠OAB=∠OBA (Isosceles triangle property)

Sum of angles of the triangle = 180

∠AOB+∠OAB+∠OBA=180

110+2∠OBA=180

∠OBA=35

∘

Since, PQ is a tangent touching the circle at B.

Thus, ∠OBQ=90

∘

Now, ∠ABQ+∠OBA=90

∠ABQ+35=90

∠ABQ=55

∘

Step-by-step explanation:

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Answer 2

Here is Your answer...

Given, ∠AOB=110 ∘

In △AOB

OA=OB (Radius of the circle)

Thus, ∠OAB=∠OBA (Isosceles triangle property)

Sum of angles of the triangle = 180

∠AOB+∠OAB+∠OBA=180

110+2∠OBA=180

∠OBA=35 ∘

Since, PQ is a tangent touching the circle at B.

Thus, ∠OBQ=90 ∘

Now, ∠ABQ+∠OBA=90

∠ABQ+35=90

∠ABQ=55 ∘

Hope it helps..❤

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