Short Note

In Fig . 10.69, the tangent at a point *C *of a circle and a diameter *AB* when extended intersect at *P* . If \[\angle\]*PCA* =110^{0}, find \[\angle\] *CBA*

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#### Solution

\[\angle ACB = 90^o \left( \text{Angle inscribed in a semi - circle} \right)\]

\[\angle PCO = 90^o \left( \text{PC is a tangent at C} \right)\]

\[Now, \angle PCA = \angle PCO + \angle OCA\]

\[ \Rightarrow 110^o = 90^o + \angle OCA\]

\[ \Rightarrow \angle OCA = 20^o\]

Since, OC = OA (radii of the circle)

\[\angle OCA = \angle OAC = 20^o\]

\[In ∆ ABC, \]

\[\angle BAC + \angle ACB + \angle CBA = 180^0\]

\[ \Rightarrow 90^o + 20^o + \angle CBA = 180^o\]

\[ \Rightarrow \angle CBA = 70^o\]

\[In ∆ ABC, \]

\[\angle BAC + \angle ACB + \angle CBA = 180^0\]

\[ \Rightarrow 90^o + 20^o + \angle CBA = 180^o\]

\[ \Rightarrow \angle CBA = 70^o\]

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