The reflex angle between the hands of a clock at 10.25 is:

1180º
2192^1º/₂
3195º
4197^1º/₂

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Angle traced by hour hand in \frac{125}{12} h r s \\ = {(\frac{360}{12} \times \frac{125}{12})}^{\circ} = 312 \frac{1^{\circ}}{2} \\ Angle traced by minute hand in 25 min \\ = {(\frac{360}{60} \times 25)}^{\circ} = 150^{\circ} \\ ∴ Reflex angle \\ = 360^{\circ} - {(312 \frac{1}{2} - 150)}^{\circ} \\ = 360^{\circ} - 162 \frac{1^{\circ}}{2} \\ = 197 \frac{1}{2} \end{aligned}

$\eqalign{ & {\text{Angle}}\,{\text{traced}}\,{\text{by}}\,{\text{hour}}\,{\text{hand}}\,{\text{in}}\,\frac{{125}}{{12}}hrs \cr & = {\left( {\frac{{360}}{{12}} \times \frac{{125}}{{12}}} \right)^ \circ } = 312\frac{{{1^ \circ }}}{2} \cr & {\text{Angle}}\,{\text{traced}}\,{\text{by}}\,{\text{minute}}\,{\text{hand}}\,{\text{in}}\,{\text{25}}\,\min \cr & = {\left( {\frac{{360}}{{60}} \times 25} \right)^ \circ } = {150^ \circ } \cr & \therefore {\text{Reflex}}\,{\text{angle}} \cr & = {360^ \circ } - {\left( {312\frac{1}{2} - 150} \right)^ \circ } \cr & = {360^ \circ } - 162\frac{{{1^ \circ }}}{2} \cr & = 197\frac{1}{2} \cr}$