Find the angle between hour-hand and minute-hand in a clock at
A) Ten past eleven
B) Twenty past seven
C) Thirty-five past one
D) Quarter to six
E) \[2:20\]
F) \[10:10\]

Hint:
Here we are asked to find the angle between hour hand and minute hand for some given values of time.
So, firstly we will find the numbers where the hour hand and minute hand are pointing and note the numbers.
Thus, we will put the value of h and m in the equation $\left| {30h - \dfrac{{11}}{2}m} \right|$ .
Hence, we get the required answers.

Complete step by step solution:
To find the angle between minute hand and hour hand we have to apply formula
 $ = \left| {30h - \dfrac{{11}}{2}m} \right|$
Where h denotes hour and m denotes minutes.
Ten past eleven means = 11:10 pm
Where \[h = 11\] , \[m = 10\]
 $ = \left| {30h - \dfrac{{11}}{2}m} \right|$
By using above formula,
Angle \[ = \left| {30 \times 11 - \dfrac{{11}}{2} \times 10} \right|\]
 \[ = \left| {330 - 55} \right|\]
 $ = {275^ \circ }$
 $ \Rightarrow $ Angle between hour hand and minute hand is ${275^ \circ }$
Twenty past seven means = 7:20
 Where \[h = 7,m = 20\]
 $ = \left| {30h - \dfrac{{11}}{2}m} \right|$ .
By using above formula,
Angle $ = \left| {30 \times 7 - \dfrac{{11}}{2} \times 20} \right|$
 $ = \left| {210 - 110} \right|$
 $ = {100^ \circ }$
 $ \Rightarrow $ Angle between hour hand and minute hand is ${100^ \circ }$
Thirty-five past one means \[ = {\text{ }}1:3\]
Where \[h = 1,m = 35\]
 $ = \left| {30h - \dfrac{{11}}{2}m} \right|$
By using above formula,
Angle $ = \left| {30 \times 1 - \dfrac{{11}}{2} \times 35} \right|$
 $ = \left| {30 - 192.5} \right|$
 $ = {162.5^ \circ }$
 $ \Rightarrow $ Angle between hour hand and minute hand is ${162.5^ \circ }$
Quarter to six means \[ = 5:45\]
Where \[h = 5,m = 45\]
 $ = \left| {30h - \dfrac{{11}}{2}m} \right|$
By using above formula,
Angle $ = \left| {30 \times 5 - \dfrac{{11}}{2} \times 45} \right|$
  $ = \left| {150 - 247.5} \right|$
 $ = {97.5^ \circ }$
 $ \Rightarrow $ Angle between hour hand and minute hand is ${97.5^ \circ }$
 \[2:20\]
Where \[h = 2,m = 20\]
   $ = \left| {30h - \dfrac{{11}}{2}m} \right|$
By using above formula,
Angle $ = \left| {30 \times 2 - \dfrac{{11}}{2} \times 20} \right|$
 $ = \left| {60 - 110} \right|$
 $ = {50^ \circ }$
 $ \Rightarrow $ Angle between hour hand and minute hand is ${50^ \circ }$
 \[10:10\]
Where \[h = 10,m = 10\]
 $ = \left| {30h - \dfrac{{11}}{2}m} \right|$
By using above formula,
Angle $ = \left| {30 \times 10 - \dfrac{{11}}{2} \times 10} \right|$
 $ = \left| {300 - 55} \right|$
 $ = {245^ \circ }$

Angle between hour hand and minute hand is ${245^ \circ }$

Note:
The above question can solve with alternate method,
Since, the above question can solve with another formula also i.e.
 \[
  \Delta \theta = \left| {{\theta _{hr}} - {\theta _{\min }}} \right| \\
  \Delta \theta = \left| {0.5^\circ \times \left( {60 \times H + M} \right) - 6^\circ \times M} \right| \\
  \Delta \theta = \left| {0.5^\circ \times \left( {60 \times H + M} \right) - 0.5^\circ \times 12 \times M} \right| \\
  \Delta \theta = \left| {0.5^\circ \times \left( {60 \times H - 11 \times M} \right)} \right| \\
 \]
Where, H is the hour and M is the minute
Since, if the angle is greater than $180^\circ $ then subtract It from $360^\circ $ .