jfan17

First, we realize that the diameter of our given circle O is 10.

That means that the length between O and the intersection with the chord is 3, because:

10-(5+2) = 10 -7 = 3

If we draw two radii to the endpoints of the chord, we get two congruent right triangles by HL congruency(their hypotenuses are both radii of the circle, and they share a common leg). The hypotenuse of one of these right triangles is 5, and their leg is 3(which we just got). To solve for the value of x, we can then write by the pythagorean theorem:

\(5^2 = 3^2 + x^2\)

\(25 = 3^2 + x^2\)

\(25 = 9 + x^2\)

\(16 = x^2\)

Square rooting both sides and discarding the negative solution(a length can't be negative), we arrive at our answer of:

\(x = 4\)

Alternatively, we can realize that the triangle created is a 3-4-5 triangle(through memorization of right triangles), which would be a faster way to calculate the length without going into pythagorean.

Just wait a bit and it'll go through moderation, I'm fairly sure

By vieta's formulas, given a cube(it can be any polynomial) ax³+bx²+cx+d = 0

and roots r,s, and t:

r+s+t = -b/a

rs +rt + st = c/a

rst = d/a

you could try solving for these values in terms of the ones that you want(for example, with 2r, 2s and 2t, rewriting these equations using those respective terms). I'm sure that this would be one approach that would work. 

If you want more information to learn more about vieta's, the internet is always a great tool to search up information regarding it.

\((x^2+y^2)^2=(x^2-y^2)^2+(2xy)^2\)

The problem gives us that:

\(x^2-y^2 =5\)

and 

\(xy =6\)

Realize that we already have all the necessary values to determine \(x^2+y^2 \) without individually solving for x or y

Substituting the values of \(x^2-y^2\) and \(xy\) that we already have:

\((x^2+y^2)^2 = (5)^2 + (2*6)^2\)

\((x^2+y^2)^2=25+(12)^2= 25 + 144 = 169\)

Now we square root both sides to get:

\(x^2+y^2 = \pm13\). The problem doesn't directly limit x and y to real numbers, so we could have an answer of -13(square rooting 169 gives a negative and positive solution). The implied answer however would be 13.

If we cube both sides of this equation(I couldn't see an easy way out of this), we get:

\(343x = x^3 + 3x^2 * 6 + 3x * 6^2 + 6^3 \)

\(343x = x^3 + 18x^2 + 108x + 216\)

Subtracting 343x on both sides, we get:

\(x^3 + 18x^2 -235x + 216 = 0\)

Realize that the sum of the coefficients is 0, which means that 1 is a root of this polynomial. Then we can factor out x-1 from this.

Doing polynomial division, we get this factored into:

\((x-1)(x^2+19x-216)\)

We factor the right hand quadratic into:

\(x^2+19x-216 = (x-8)(x+27)\)

We have then factored our cubic into:

\((x-1)(x-8)(x+27) = 0\)

This gives us our three solutions(by the fundamental theorem of algebra, there should be 3), as :
 

1, 8, and -27

We can make use of the power of a point theorem, a useful application for geometry. 

It has 3 variations, and in this case, I'll use 1 of them : the case with two secants intersecting a circle(in other words, the lines intersect the circle at two points each).

It states:

Given the two secants in this problem GJ and GL,

GH * GJ = GK * GL

we are given that:

GH = 13

GJ = z + 13

GK = 9

GL = 39

Substituting, we get:

13(z+13) = 9 * 39

Dividing by 13 on both sides, we get:

z+ 13 = 9 * 3

z + 13 = 27

z = 14

secant = 1/cos

Dividing by -sqrt3 on both sides

\(\sec{\theta} = -\frac2{\sqrt3}\)

Rationalizing the denominator by multiplying it by sqrt3/sqrt3(which is one):

\(\sec{\theta} = -2\sqrt{3}/3\)

Substituting in 1/cos:

\(1/\cos{\theta} = -2\sqrt{3}/3\)

\(\cos{\theta} = -3/2\sqrt{3}\)

Rationalizing like before by multiplying by sqrt3 / sqrt 3, we get:

\(\cos{\theta} = -3\sqrt{3}/6 = -\sqrt3/2\)

The values for which this is true between [0, 2pi] are :

\(\cos{5\pi/6}\) radians (cos 150 degrees) and \(\cos{7\pi/6} \)(cos 210 degrees). The general forms for these are then indeed your answer because the period of cosine is 2pi.

No, we in fact don't include 0 or pi. What's your reasoning for that?(Just curious).

The cosine of 0 is equal to 1. However,  

2* 1^2 is not equal to 1, meaning that 0 does not follow through with our given equation.

Let's check \(\pi\)

the cosine of \(\pi\) = -1. Similarly, 

2 * (-1) ^2 is not equal to -1, meaning that pi does not follow through with our given equation either.

For me it says that the picture is waiting for moderation

We can use a theorem in geometry known as "heron's formula". It states that given semi-perimeter(half of the perimeter) of triangle ABC as s, we have:

[ABC] = \(\sqrt{s(s-a)(s-b)(s-c)}\)

The semi perimeter of this triangle is:

\({13+15+24\over2} = {52\over2} = 26\)

We then have:

[ABC] = \(\sqrt{26(26-13)(26-15)(26-24)} = \sqrt{26*13*11*2} = \sqrt{26*26*11}\)

we can then take out a 26 from the square term, which gets us:

\(26\sqrt{11}\) in² as our area