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प्रश्न
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.
उत्तर
In the given figure,
\[ \Rightarrow \angle AOC = 180^o - 130^o = 50^o\]
\[ \Rightarrow \angle AOC = 180^0- 130^o = 50^0\] (Because the tangent at any point of a circle is perpendicular to the radius through the point of contact)
\[\angle AOC + \angle CAO + \angle ACO = 180^o \left( \text{Angle sum property of a triangle} \right)\]
\[\angle AOC + \angle CAO + \angle ACO = 180^o\left( \text{Angle sum property of a triangle} \right)\]
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