CPhill

I believe "Anonymous" made a slight mistake, here.

The +8x in the answer should be -8x

Very interesting what Melody said......there is a book by Petr Beckmann entitled "A History of Pi". I don't know if it is still in print, or not, but it is quite fascinating. However, the last half of the book requires some fair knowledge of upper maths!!

One funny note....Many years ago, The State Legislature of Indiana in the US once tried to "legislate" the value of pi!!! Go figure....

Just divide the numerator by the denominator....that will give you your decimal equivalent.

Using the arcsin function, we have

arcsin(3/7.6)  ≈ 23.25°

Realize that this angle might also lie in the 2nd quadrant, as well.

There, the angle would be ≈ 156.75°

Let's see if we can solve this......

In the second equation, we can add y to both sides and subtract 2 from both sides.

So we have

y = x^2 - 2     and

y = -3x + 2     So setting the "y's" equal, we have

x^2 - 2 = -3x + 2     Subtracting everything on the right on both sides, we have

x^2 + 3x - 4 = 0       And factoring, we have

(x + 4) (x - 1) = 0     And setting both factors to 0, we have

x = -4 and x = 1        And using y = -3x + 2 and substituting both values of x

We find that y = 14 when x = -4 and y =  -1 when x = 1

So our solutions are  (-4, 14)  and ( 1, -1)

Yeah, Alan....I realized that I could have "manipulated" the Geogebra generated object to produce any number of quads with these sides....this one just had a right angle which was fairly easy to deal with.

....But, at the moment.....I'm still looking for a missing zero along with my pal Sisyphus.....no more time for this "four-sided" silliness!!

Let's convert the yards and feet to inches

7 yards = 7 * 36 = 252 inches     2 feet = 24 inches

So, I'm assuming that the pot is shaped like a cylinder?? If so, the "formula for the volume of a cylinder is

V = pi * r^2 * h         and since the diameter is 252 inches, the radius = 126 inches

So, using 3.14 for "pi," we have

V = 3.14 * (152 in)^2 * (24) =  1741117.44 in³

and the volume of each ice cube is 2in x 2in x 2in = 8 in³

And dividing the pot's volume by the volume of each ice cube we have

1741117.44 in³ / 8 in³ = 217639.68 ≈ 217640 ice cubes   

You'd better hurry, though before they melt.....then, we'll have to figure how many gallons of water the pot holds!!

We need to use the sin for this

sin θ = opp / hyp

sin 42 = h / 33     where h is the height of the kite

33*sin 42 = h = 22.08 ft

To get the length of the other diagonal, DB, we'll use this "formula" to find the measure of angles DAC and CAB :

Cos θ  =   [u(dot)v] /( llull  * llvll)          Where u and v are vectors and

[u(dot)v] - the dot product of u and v  and

llull = length of u and

llvll = length of v

So, taking AD and AC as vectors

AD =<5.4, 39.7>  AC = <40.6, 40.2>    AD(dot)AC = 1815.18

llADll = 40    llACll = 57.1

cos^-1 (1815.18)/(40*57.1) = m< DAC =37.37

And again, taking AB and AC as vectors

AB = <40.6 ,0>  AC <40.6, 42.7>          AB(dot)AC = 1648.36

llABll = 40.6    llACll = 57.1     

cos^-1 (1648.36)/(40.6*57.1) = m< CAB = 43.8

m< DAB = m< DAC + m< CAB = 37.37 + 44.7 = 82.07

And using the Law of Cosines, DB = SQRT (40^2 + 40.6^2 - 2(40)(40.6)cos(82.07)) = 52.92 ≈ 53

And that's close enough for me!!

Well ...IF you could see the whole thing (I'm tired of messing with it!!) The sides AB and BC (BC is off the chart, so to speak) form a right angle

So sqrt(40.6^2 + 40.2^2) =57.1 and that's the length of AC

I haven't actually figured out a way to calculate the other diagonal...if there is one?? I might look at it later and see if anything "pops" into my head.  Maybe we could set it up as a question in the "puzzles" section??

Besisdes, the questioner only asked for "a" diagonal.......he didn't atually specify WHICH one !!

I claim I've actually done my duty!!!

Time for lunch!!