Question

need quality answer
in figure ray bc is a tangent at point b and ray ba is secant .
angle ABC interupts arc aXb
if m arc aXb = 130° then find measure of angle ABC

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Answer 1

Answer:

$\huge \fbox \colorbox{lime}{Solution ✍}$

$\tt \pink{Angle \: ABC=65° }\: is \: the \: ans$

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Step-by-step explanation:

refer the attachment️‍️️‍️

Jo in the po.in.t A and B by cen ter O.

So, the angle AOB = 130°

Let the angle ABC = θ

OA = OB

Since, OB are the radius .

Therefore , angleOBA = angle OAB = x° _____eq(1)

( equal sides have equal angle op posite to it)

Now,

In the figure,

BC is the tangent to circle at po.intt B

And , measure of a( AXB) = 130°

therefore, angle BOA = 130° ( by arc theorem)

Since , in triangle , AOB

OA = OB (radius)

angle OBA = angle OAB = x°. (From eq1)

In triangle AOB

angle O + angle A + angle B = 180°

=> 130° + 2x° = 180°

=> 2x = 50°

=> x = 25°

Now,

By tangent theorem,

Radius will be perpendicular to tangent at po.intt of con.tact , angle OBC = 90°

angle OBA + angle ABC = 90°

=> x° + θ= 90°

=> 25° + θ= 90°

=>θ = 65°

Therefore , angle ABC = 65°.

Answer 2

Step-by-step explanation:

️‍️

️‍️️‍️

Step-by-step explanation:

refer the attachment️‍️️‍️

Jo in the po.in.t A and B by cen ter O.

So, the angle AOB = 130°

Let the angle ABC = θ

OA = OB

Since, OB are the radius .

Therefore , angleOBA = angle OAB = x° _____eq(1)

( equal sides have equal angle op posite to it)

Now,

In the figure,

BC is the tangent to circle at po.intt B

And , measure of a( AXB) = 130°

therefore, angle BOA = 130° ( by arc theorem)

Since , in triangle , AOB

OA = OB (radius)

angle OBA = angle OAB = x°. (From eq1)

In triangle AOB

angle O + angle A + angle B = 180°

=> 130° + 2x° = 180°

=> 2x = 50°

=> x = 25°

Now,

By tangent theorem,

Radius will be perpendicular to tangent at po.intt of con.tact , angle OBC = 90°

angle OBA + angle ABC = 90°

=> x° + θ= 90°

=> 25° + θ= 90°

=>θ = 65°

Therefore , angle ABC = 65°.

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