Finding angles of a parallelogram.

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Homework Help Overview

The discussion revolves around finding the angles of a parallelogram given its area and the lengths of its diagonals. Participants are exploring methods to derive a solution that applies to non-rhombus shapes.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss using the area formula and relationships involving height and angles. Some mention the Law of Cosines in relation to the diagonals, while others express confusion about deriving side lengths without right angles. Questions arise regarding the necessity of the angle of intersection for finding sides.

Discussion Status

The discussion is active, with various approaches being considered. Some participants have proposed solutions using trigonometric relationships and the Law of Cosines, while others are questioning the assumptions and information needed to proceed. There is no explicit consensus on a single method yet.

Contextual Notes

Participants note the challenge of working with a non-rhombus parallelogram and the implications of missing information, such as the angle of intersection of the diagonals. The constraints of the problem are acknowledged, particularly in relation to the application of trigonometric principles.

tiggertime
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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
 
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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).
 
HallsofIvy said:
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.
 
I got a solution using the Law of Cosines with the angle of intersection of the diagonals to get the lengths of the sides.
 
insightful said:
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.
 
tiggertime said:
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.
 
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