# Finding angles of a parallelogram.

Tags:
1. Oct 2, 2015

### tiggertime

1. The problem statement, all variables and given/known data
This question has popped up recently and I was completely stumped. How to find the angles of a parallelogram given only area and the length of the diagonals? I'm trying to find a generic solution or formula that works for a non-rhombus.

2. Relevant equations

3. The attempt at a solution

Parallelogram and Triangle Laws
Setting angle variable equal height(sin(angle variable))

2. Oct 2, 2015

### HallsofIvy

Staff Emeritus
The area of a parallelogram is "base times height". "Base" is the length of one side, b. The height is the length of perpendicular from a vertex to the base. Given angle $\theta$ the height is the "opposite side" while the length of that side, a, is the hypotenuse, h, so $h= a sin(\theta)$ The area is $ab sin(\theta)$.

3. Oct 3, 2015

### tiggertime

I still don't understand how to obtain sides a and b from the diagonal lengths since there are no right angles to use Pythagoras.

4. Oct 3, 2015

### insightful

I got a solution using the Law of Cosines with the angle of intersection of the diagonals to get the lengths of the sides.

5. Oct 3, 2015

### tiggertime

So you were given the angle of intersection? How do you find the sides without that information? I tried substituting A=absinθ into the equation, after working through sine changes, but the solution keeps cancelling out.

6. Oct 3, 2015

### insightful

No, you can use some simple trig in one quadrant of the parallelogram to get the angle (in terms of A and diagonals d1 and d2). Remember, each quadrant has 1/4 the total area A.

Last edited: Oct 3, 2015