Question 1134517
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            I will give you only basic instructions, determining the direction and leaving the details to you.



<pre>
1.  Let O be the center of the circle circumscribed around the triangle ABC.

    Then the angle BOC is 60 degrees.

    The triangle BOC is ISOSCELES with two sides of 24 cm long (the radii) and the angle BOC of 60 degrees.

     Use the Cosine Law to find BC.



2.  As soon as you find BC, you know BM as half of BC.



3.  Then the triangle BMP is a right angled triangle, in which you know the legs.

    Then find the hypotenuse PB, which is what you need to find.
</pre>


Happy calculations !



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<U>Later addition</U>


<pre>
The triangle  BOC is ISOSCELES  with the angle  BOC  of 60 degrees between two equal sides ===============>


    hence, it is EQUILATERAL, and all its sides are 24 cm long.


So,  BC  is 24 cm;  hence,  BM is 12 cm (as half of BC).


Thus  |PB| = {{{sqrt(12^2 + 7^2)}}} = {{{sqrt(144 + 49)}}} = {{{sqrt(193)}}}.


<U>ANSWER</U>.  |PB| = {{{sqrt(193)}}} cm = 13.89 cm (approximately).
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Solved.



Thus I solved it &nbsp;<U>COMPLETELY</U> &nbsp;without leaving the room for you &nbsp;(for your calculations . . . ).  &nbsp;Sorry for it . . .