Given that AB and AC are tangents to the circle and AC = BC then angle CAB!? See attachment please.
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Answered by
2
Step-by-step explanation:
OC⊥DA ∴ ∠OCD=90∘
In △OCD,∠ODC=180−∠OCD−∠DOC
∠ODC=180∘−(90∘+65∘)
∠ODC=180∘−155∘=25∘
Also OB⊥AB, ∴ ∠OBA=90∘
In △DBA,∠DAB=180∘−(∠BDA+∠DBA)
=180∘−(25∘+90∘)
=180∘−115∘=65∘
∴ ∠OAB=21×∠CAB(or ∠DAB)=21×65∘=32.5∘
(∴ Tangents are equally inclined to the line joining the external pt. and the centre of the circle)
In △OBA,X=180−∠OAB−∠OBA=180∘−(90
Answered by
8
Given:-
- AB and AC are tangents to the circle.
- AB=BC
To find:-
- <CAB
Answer:-
We know that
Tangents to a circle from an external point of the circle are equal.
And Given that
- From eq(1) and eq(2)
Hence its an equilateral triangle.
- There fore its all angles are equal.
Let each be x
- We know that in a triangle
Therefore
- <CAB=60°
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