19. In the given figure, the tangent at a point C of a circle and a diameter AB when extended intersect at P.
If ZPCA = 130°, find ZCBA.
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Answer:
Let O be the center of the circle.
A,O,B,P all are on the same line and P and C are points on the tangent.
AB is a diameter of a circle.
∴ ∠BCA=90
o
[ Angle inscribe in a semi-circle. ]
C is the point on the circle where the tangent touches the circle.
⇒ So, ∠OCP=90
o
.
⇒ ∠PCA=∠PCO+∠OCA
⇒ 110
o
=90
o
+∠OCA
⇒ ∠OCA=20
o
In △AOC,
⇒ AO=OC [ Radius of a circle. ]
⇒ ∠OCA=∠CAO=20
o
Step-by-step explanation:
In △AOC,
⇒ AO=OC [ Radius of a circle. ]
⇒ ∠OCA=∠CAO=20
o
In △ABC,
⇒ ∠CAB+∠CBA+∠BCA=180
o
⇒ 20
o
+∠CBA+90
o
=180
o
⇒ 110
o
+∠CBA=180
o
∴ ∠CBA=70
o
.
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