Question

In Fig. 1, AOB is a diameter of a circle with centre O and AC is a tangent to the circle at A. If ∠BOC = 130°, the find ∠ACO.

Answer 1

In the given figure, 

\[\angle BOC = 130^o\] 

\[\angle BOC = 130^o\] form a linear pair.

So,

\[\angle BOC + \angle AOC = 180^o\]
\[ \Rightarrow \angle AOC = 180^o - 130^o = 50^o\]  

\[\angle BOC + \angle AOC = 180^o\]
\[ \Rightarrow \angle AOC = 180^0- 130^o = 50^0\]                                         (Because the tangent at any point of a circle is perpendicular to the radius through the point of contact)                            

\[In ∆ AOC, \]
\[\angle AOC + \angle CAO + \angle ACO = 180^o \left( \text{Angle sum property of a triangle} \right)\]

\[In ∆ AOC, \]
\[\angle AOC + \angle CAO + \angle ACO = 180^o\left( \text{Angle sum property of a triangle} \right)\]

\[\angle ACO = 40^o\]