In figure, AB is a chord of the circle and AOC is its diameter such that `angleACB=50^(@)`. If AT is the tangent to the circle at the point A, then `angleBAT` is equal to

Options

A. `45^(@)`
B. `60^(@)`
C. `50^(@)`
D. `55^(@)`

Answer

C

Explanation

In figure, AOC is a diameter of the circle. We know that, diameter subtends `90^(@)` at the circle.
So, `angleABC=90^(@)`
In `DeltaACB, angleA+angleB+angleC=180^(@)`
[since, sum of all angles of a triangle is `180^(@)`
`rArrangleA+90^(@)+50^(@)=180^(@)`
`rArrangleA+140=180`
`rArrangle=180^(@)-140^(@)=40^(@)`
`angleAorangleOAB=40^(@)`
Now, AT is the tengent to the circle at point A. So, OA is perpendicular to AT.
`:.angleOAT=90^(@)` [from figure]
`rArrangleOAB+angleBAT=90^(@)`
On putting `angleOAB=40^(@),` we get
`rArrangleOAB+angleBAT=90^(@)`
Hence, the value of `angleBAT` is `50^(@)`