Question 1207171
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What is the distance between the tips of the minute hand and the hour hand 
of a clock at 1:35pm where the length of the minute hand is 14cm 
and the length of the hour hand is 9cm
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<pre>
At 1:35 pm, the angle between the minute hand and vertical direction up is  

    {{{(35/60)*2pi}}} = {{{(35/30)pi}}} = {{{(7/6)pi}}}  radians

(since the minute hand makes a full rotation in 60 minutes).


The angle between the hour hand and vertical direction up is

     {{{(1/12)*2pi}}} + {{{(35/(12*60))*2pi}}} = {{{(1/6)pi}}} + {{{(7/72)pi}}} = {{{(19/72)pi}}}.

(since the hour hand makes a full rotation in 12 hours = 12*60 minutes).


The difference between these angles is

    {{{(7/6)pi}}} - {{{(19/72)pi}}} = {{{(84/72)pi}}} - {{{(19/72)pi}}} = {{{(65/72)pi}}}.


Thus the angle between the two hands at 1:35 pm is  {{{(65/72)pi}}}  radians.

It is more than the right angle {{{pi/2}}} = {{{(36/72)pi}}}, but less than the straight angle of {{{pi}}} = {{{(72/72)pi}}}.



Now, we have an obtuse triangle with the sides of 14 cm and 9 cm and the angle of {{{(65/72)pi}}}  radians between them.


To find the distance between the tips of the hands, apply the cosine law equation

    d = {{{sqrt(14^2+9^2-2*14*9*cos((65/72)*3.14159265))}}} = 22.745 cm.


<U>ANSWER</U>.  The distance between the tips is about  22.745 cm.
</pre>

Solved.