CIRCLE
to a circle are equal and
Radius is perpendicular to the tangent
A
P
B
O
?
In ◻APBO,
Sol:
∠PAO
=
∠PBO
=
90º
[Radius is perpendicular
to the tangent]
∠AOB
+
∠APB
+
360º
∠PAO
+
∠PBO
=
[Sum of all angles of quadrilateral is 360º]
∴
∠AOB
+
80º
+
360º
90º
+
90º
=
∴
=
260
360
∠AOB
+
∴
100º
∠AOB
=
∠APB
=
[Given]
80º
…(i)
Q. If tangents PA and PB from a point P to a circle with centre O
are inclined to each other at angle of 80°, Find ∠POA
80°
We know, sum of all angles of quadrilateral is 360º
∴
–
360
260
∠AOB
=
Consider □APBO
We know, radius is perpendicular to tangent
∠PAO = 90º
Observe ∠PAO
∠PBO = 90º
Observe ∠PBO
A
P
B
O
PA = PB
[Length of the tangents drawn from an
external point to a circle are equal]
OA = OB
[Radii of the same circle]
OP = OP
[Common side]
∴
ΔPOB
ΔPOA
≅
[SSS test]
∠AOB
=
∠POA
+
∠POB
…(iii)
[Angle addition property]
∴
100
=
2∠POA
∴
∠POA
=
50o
1000
∠AOB
=
In ΔAOP and ΔBOP
∴
∠POB
∠POA
=
…(ii)
[c.p.c.t]
∴
Sol:
Q. If tangents PA and PB from a point P to a circle with centre O
are inclined to each other at angle of 80°, Find ∠POA
∠AOB is made up of two angles
∠POA,
∠POB
Consider ΔAOP and ΔBOP
?
∠AOB
=
∠POA
+
∠POA
∴