If the angle between two radii of a circle is 130°, then the angle between the tangent at the ends of the radii is
(a) 65°
(b) 40°
(c) 50°
(d) 90°
(c) $50^{\circ}$
$\mathrm{OA}$ and $\mathrm{OB} a$ re the two $\mathrm{r}$ adii of $a$ circle with centre $\mathrm{O}$.
Also, $A P$ and $B P$ are the tangents to the circle.
Given, $\angle A O B=130^{\circ}$
Now, $\angle O A B=\angle O B A=90^{\circ}$ (S ince tangents drawn from an external point a re
perpendicular to the radius at point of contact)
In quadrilateral $O A P B$,
$\Rightarrow 130^{0}+90^{0}+90^{0}+\angle A P B=360^{0}$
$\Rightarrow \angle A P B=360^{0}-\left(130^{0}+90^{0}+90^{0}\right)$
$\Rightarrow \angle A P B=360^{0}-310^{\circ}$
$\Rightarrow \angle A P B=50^{\circ}$
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