Question

please answer!!!

no waste answers please​

Answer 1

Answer:

in given figure, Triangle OAB

First find <OAB=90-50=40 [ tangent and radios makes perpendicular ]

since OA=OB = radios

so 40+40+<OAB=180

<OAB= 180-80=100

Answer 2

Answer:

∠AOB is 100°.

Step-by-step explanation:

To find :

• Measure of ∠AOB.

Solution :

Construction : Draw a line BC, which will intersect line AP at point C and BC will be tangent to circle with centre O at point B.

We know theorem :

✿ The lengths of the two tangents from an external point to a circle are equal.

So, Length of tangent AC = Length of tangent BC.

In ∆ABC :

If two sides of triangle are equal then their opposite angles are also equal. So, Side BC and AC are equal then,

∠BAC (or BAP) = ∠ABC = 50°.

• We have to find measure of ∠AOB. So,

We also know theorem :

✿ The tangent to a circle is perpendicular to the radius through the point of contact.

So, ∠OAC = 90° and ∠OBC = 90°.

In ∆AOB :

• ∠OAB = ∠OAC - ∠BAC = 90° - 50° = 40°.

• ∠OBA = ∠OBC - ∠ABC = 90° - 50° = 40°.

Now,

By angle sum property of triangle :

$\implies$ ∠AOB + ∠OAB + ∠OBA = 180°

$\implies$ ∠AOB + 40° + 40° = 180°

$\implies$ ∠AOB + 80° = 180°

$\implies$ ∠AOB = 180° - 80°

$\implies$ ∠AOB = 100° .

Therefore,

Measure of ∠AOB is 100°.