Question

If the angle between two radii of a circle is 130°, then the angle between the angles at the ends of the radii is
Figure

(a) 65°
(b) 40°
(c) 50°
(d) 90°

Answer 1

$$\begin{gathered} \left(c\right)50° \\ \mathrm{OA}\mathrm{and}\mathrm{OB}a\mathrm{re}\mathrm{the}\mathrm{two}\text{r}\mathrm{adii}\mathrm{of}a\mathrm{circle}\mathrm{with}\mathrm{centre}\mathrm{O.} \\ Also,\mathit{AP}\mathit{}\mathrm{and}\mathit{}\mathit{BP}\mathit{}\mathrm{are}\mathrm{the}\mathrm{tangents}\mathrm{to}\mathrm{the}\mathrm{c}\mathrm{ircle}. \\ \mathrm{Given},\angle \mathit{AOB}={130}^0 \\ \mathrm{Now},\angle \mathit{OAB}=\angle \mathit{OBA}={90}^0(\text{S}\mathrm{ince}\mathrm{tangents}\mathrm{drawn}\mathrm{from}\mathrm{an}\mathrm{external}\mathrm{point}\mathrm{a}\mathrm{re} \\ \mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{radius}\mathrm{at}\mathrm{p}\mathrm{oint}\mathrm{of}\mathrm{c}\mathrm{ontact}\mathit{)} \\ \mathrm{In}\mathrm{quadrilateral}\mathit{}\mathit{OAPB}\mathit{,} \\ \angle \mathit{AOB}+\angle \mathit{OAB}+\angle \mathit{OBA}+\angle \mathit{APB}\mathit{=}{360}^0 \\ \mathit{\Rightarrow }{130}^0+{90}^0+{90}^0+\angle \mathit{APB}\mathit{=}{360}^0 \\ \mathit{\Rightarrow }\angle \mathit{APB}={360}^0-\left({130}^0+{90}^0+{90}^0\right) \\ \mathit{\Rightarrow }\angle \mathit{APB}={360}^0-{310}^0 \\ \mathit{\Rightarrow }\angle \mathit{APB}={50}^0 \end{gathered}$$