in the given figure ,AB is a chord of a circle and PQis tangent at point B of the circle if angle AOB equal to 110 0 then angle ABQ is
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Answered by
10
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°
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Answered by
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Answer:
Step-by-step explanation:
Solution
Given,
∠
A
O
B
=
110
∘
In
△
A
O
B
O
A
=
O
B
(Radius of the circle)
Thus,
∠
O
A
B
=
∠
O
B
A
(Isosceles triangle property)
Sum of angles of the triangle = 180
∠
A
O
B
+
∠
O
A
B
+
∠
O
B
A
=
180
110
+
2
∠
O
B
A
=
180
∠
O
B
A
=
35
∘
Since, PQ is a tangent touching the circle at B.
Thus,
∠
O
B
Q
=
90
∘
Now,
∠
A
B
Q
+
∠
O
B
A
=
90
∠
A
B
Q
+
35
=
90
∠
A
B
Q
=
55
∘
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