Question

Given that AB and AC are tangents to the circle and AC = BC then angle CAB!? See attachment please.

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Answer 1

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

Answer 2

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.

$\\ \sf\longmapsto AB=AC\dots(1)$

And Given that

$\\ \sf\longmapsto AB=BC\dots(2)$

• From eq(1) and eq(2)

$\\ \sf\longmapsto AB=BC=AC$

Hence its an equilateral triangle.

• There fore its all angles are equal.

Let each be x

• We know that in a triangle

$\\ \sf\longmapsto Sum \:of\:angles=180$

$\\ \sf\longmapsto x+x+x=180$

$\\ \sf\longmapsto 3x=180$

$\\ \sf\longmapsto x=\dfrac{180}{3}$

$\\ \sf\longmapsto x=60°$

Therefore

• <CAB=60°