Finding angles of a parallelogram.

[tiggertime]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

[/B]
Parallelogram and Triangle Laws
Setting angle variable equal height(sin(angle variable))
Vector addition laws

 

[HallsofIvy]

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).

 

[tiggertime]

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).

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.

 

[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.

 

[tiggertime]

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

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.

 

[insightful]

So you were given the angle of intersection?

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.