Question

AB ← → is tangent to circle O at point A. If m∠AOB = 55°, what is m∠ABO ?

Answer 1

`m\angle ABO`=
`35^(\circ)`

Explanation:

Given AB is a tangent to the circle O and touch the circle at point A.

We know that tangent is a line which touches the circle only at one point .

`m\angle AOB=55^(\circ)`

We know that the radius of circle is perpendicular to the tangent.

In the figure OA is radius of given circle

AB is a tangent line

Radius OA is perpendicular to the tangent AB.

Therefore,
`m\angle OAB=90^(\circ)`

We know that sum of angles of a triangle is
`180^(\circ)`.

By sum of angle of triangle property

Therefore,in triangle OAB

`m\angle`OAB+
`m\angle`ABO+
`m\angle`AOB=
`180^(\circ)`

By angle sum of triangle property

90+55+
`m\angle ABO=180`

`145+m\angle ABO=180`

`m\angle ABO=180-145=35^(\circ)`

Hence, the angle ABO=
`35^(\circ)`.

Answer 2

A tangent forms a right angle with the centre of circle at the point it touches the circumference. i.e. OAB = 90 degrees
If AOB = 55 degrees, then ABO = 90 - 55 = 35 degrees