Question

14.
In the figure AB and AC are the two tangents drawn from the point A to the
circle with centre O. If | BOC = 130° then find | BAC
B
Q130°
А
C​

Answer 1

Answer:

180-130=50....hope helps u

Answer 2

We know that the tangent is always  perpendicular to the radius.

Therefore, we can say that the quadrilateral formed with ABOCA is always equal to $360^o$.

and given $BOC=130^o$

and  we also know the angles between radius and tangent equal to $90^o$

By calculating, we get:

$360^o=90^o+130^o+90^0+BAC$

$BAC=360^o-90^o-90^o-130^o$

$BAC=50^o$.