# In Fig . 10.69, the Tangent at a Point C of a Circle and a Diameter Ab When Extended Intersect at P . If ∠ Pca =1100, Find ∠ Cba - Mathematics

Short Note

In Fig . 10.69, the tangent at a point of a circle and a diameter AB when extended intersect at P . If $\angle$PCA =1100, find $\angle$ CBA

#### 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)

$\angle OCA = \angle OAC = 20^o$
$In ∆ ABC,$
$\angle BAC + \angle ACB + \angle CBA = 180^0$
$\Rightarrow 90^o + 20^o + \angle CBA = 180^o$
$\Rightarrow \angle CBA = 70^o$
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#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 8 Circles
Exercise 8.2 | Q 40 | Page 39