Melody

Oh, you cannot use Heron's Rule.

that is for finding the area when you have the sides.

OR

somtimes you may be able to use it fo find a side when given an area 

BUT this question has nothing to do with area :))

Manas has asked me to explain my answer more fully. (by private message)

I drew up the diagram. Let X be the centre of the cirlce.  (I fogot to put that on the diagram) 

Draw in each radii and draw in t he intervals between the centre ifthe circle and the vertices

The angle between a radius and a tangent is 90 degrees 

So you have now split your triangle up into 6 right angled triangles

cr=cp=8   tangents subtended from a point intersect the circle at equal distances.

br=bQ=6

ap=aq = ?

\(tan\beta = 4/6\\ so\\ \beta = tan^{-1}(4/6) = 33.6900675\; degrees \)

\(tan\delta=4/8\\ \delta=tan^{-1}(4/8)=26.5650511\; degrees\)

\(2\alpha+2\beta+2\delta=180\;\;angle\;sum\;of\;a\;triangle\\ \alpha+\beta+\delta=90\\ \alpha=90-\beta-\delta\\ \alpha=90-33.6900675-26.5650511\\ \alpha=29.74488\\ tan\;\alpha=\frac{4}{distance\;Aq}\\ distance\;Aq=\frac{4}{tan\;29.74488}\\ distance\;Aq=\frac{4}{tan\;29.74488}\\ distance\; Aq\approx6.99888\\ Aq=Ap\approx 7\;units \)

If you still do not understand manas you can say so but you need to ask specific questions. 

If you do not understand any of it then perhaps it is just way over your head. :)

Here is an example.  You can just type it into the input box on the calc if you want.   :))

7+sqrt(3/4)=

Thanks Chris :D

@@ What is Happening?  [Wrap4]    Wed  21/10/15     Port Macquarie, Australia Time 11:30pm   ♪ ♫

Hi,

This is not much of a wrap I am afraid, I have been travelling today and there are many questions that i have not seen, and also quite a few that do not have asnwers.  I give a big thank you to our diligent answerers and I am sorry if your question has not been answered.  Perhaps it still may be answered,  but you can repost if your question if it is no longer on the first page.  It would however, be good if you said it was a repost and perhaps give the address of the original post. :)

Here are a few interesting questions/answers.  

1) Since back the the future has, at one point, today been the present.     (Wed 21/10 2015)

    I feel I must include this question.  It is at present one of the many that has NOT been answered. :)

    http://web2.0calc.com/questions/this-question-is-for-back-to-the-future-day

2) Inscribed circle - finding lengths.          Thanks Melody and Chris

    This is a repeat question.  Chris drew a great diagram and he did it a little differently from me.  

     http://web2.0calc.com/questions/geomerty-sum#r1

3) Partial Fractions       Melody

   http://web2.0calc.com/questions/partial-fraction-decomposition_1

4) Limit question   Thanks CPhill

  http://web2.0calc.com/questions/limit-question#r2

If you want to comment an any issues please do so.    

Enjoy the rest of your day :) 

Melody    ♪ ♫

----------------------------------------------------

Lantern thread:

http://web2.0calc.com/questions/lantern-post-wed-21-10-15-back-to-the-future-day-d

Very impressive Chris     

Hi, I think ths is what you are after :))

\(\frac{13x+2}{(x-1)^2+x+1}\\ =\frac{13x+2}{x^2-2x+1+x+1}\\ =\frac{13x+2}{x^2-x+2}\\ =\frac{13x+2}{(x-2)(x+1)}\\ =\frac{A}{(x-2)}+\frac{B}{(x+1)}\qquad Where\;\;A \;and\;B\; are\; real\;numbers.\\ =\frac{A(x+1)+B(x-2)}{(x-2)(x+1)}\\ \)

\(\qquad so\\ \qquad Ax+A+Bx-2B=13x+2\\ \qquad (A+B)x+(A-2B)=13X+2\\ \qquad A+B=13\;\;(1) \qquad A-2B=2\;\;(2)\\ \qquad (1)-(2)\\ \qquad 3B=11\\ \qquad B=11/3\\ \qquad Substitution\;gives\; A=28/3\\ \)

\(=\frac{28}{3(x-2)}+\frac{11}{3(x+1)}\)

I am not sure what this question even means Siaff. 

Am I right in assuming you already know the answer to your intended question?  

20 x 52 cents

Haha Chris.

I answered first.