At 3:40, the hour hand and the minute hand of a clock form an angle of:

1120°
2125°
3130°
4135°

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Angle traced by hour hand in 12 hrs = 36 0^{\circ} \\ Angle traced by it in \frac{11}{3} hrs \\ = {(\frac{360}{12} \times \frac{11}{3})}^{\circ} = 110^{\circ} \\ Angle traced by min . hand in 60 min = 36 0^{\circ} \\ Angle traced by it in 40 min . \\ = {(\frac{360}{60} \times 40)}^{\circ} = 240^{\circ} \\ ∴ Required angle \\ = {(240 - 110)}^{\circ} = 130^{\circ} \end{aligned}

$\eqalign{ & {\text{Angle}}\,{\text{traced}}\,{\text{by}}\,{\text{hour}}\,{\text{hand}}\,{\text{in}}\,12\,{\text{hrs}}\,{\text{ = }}\,{\text{36}}{{\text{0}}^ \circ } \cr & {\text{Angle}}\,{\text{traced}}\,{\text{by}}\,{\text{it}}\,{\text{in}}\,\frac{{11}}{3}\,{\text{hrs}} \cr & = {\left( {\frac{{360}}{{12}} \times \frac{{11}}{3}} \right)^ \circ } = {110^ \circ } \cr & {\text{Angle}}\,{\text{traced}}\,{\text{by}}\,{\text{min}}{\text{.}}\,{\text{hand}}\,{\text{in}}\,60\,{\text{min}}\,{\text{ = }}\,{\text{36}}{{\text{0}}^ \circ } \cr & {\text{Angle}}\,{\text{traced}}\,{\text{by}}\,{\text{it}}\,{\text{in}}\,{\text{40}}\,{\text{min}}. \cr & = {\left( {\frac{{360}}{{60}} \times 40} \right)^ \circ } = {240^ \circ } \cr & \therefore {\text{Required}}\,{\text{angle}} \cr & = {\left( {240 - 110} \right)^ \circ } = {130^ \circ } \cr}$