I need to find the value of x in a rhombus.

Question

I need to find the value of x in a rhombus.  The equations I'm given are for half of a corner angle on the top right corner and half of a corner angle on the top left corner.

 

 

Answer 1

The angles you gave me aren't opposing, correct. I did not say that they were. I was meaning in ∠ABCD, A and C are opposite and equal, as are B and D.

 

Now that I have the equations, I can help you further.

What you need is to move the 420 back to the other side. That way, you have a quadratic formula:

16x² + 4x - 420 = 0

We can factor out a 4 from the equation and then divide both sides by 4:

4(4x² + x - 105) = 0

4x² + x - 105 = 0

 

That 4 in front gives us several possibilities to waste time with factoring, so let's try the quadratic formula:

x = (-b ± √(b² - 4ac)) / 2a

x = (-1 ± √1² - 4 * 4 * -105)) / 2 * 4

x = (-1 ± √1² + 1680)) / 8

x = (-1 ± √1681)) / 8

x = (-1 ± 41) / 8

x = (-1 + 41) / 8 Since we can't have a negative degree

x = 40/8

x = 5

Answer 2

16x^2+4x=420

16x^2+4x-420=0

divide both sides by 4

4x^2+x-105=0

factor

4*105=2*2*3*5*7=(2*2*5)*(3*7)=20*21 and 21-20=1, the coefficient of x

(4x     )(x     )=0 so far

(4x   21)(x   5)=0 so far

(4x+21)(x-5)=0

x-5=0 or 4x+21=0 but x can't be negative because we are talking about interior angles of a rhombus

x-5=0

x=5

your angles, according to your expressions, are 60º and 30º (these are half-angles)

the 4 angles are 120, 120, 60, and 60