Solve Arrow Head Geometry Question with L1, L2 & L3

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In summary, the conversation is about solving for unknown angles and lengths in a non-symmetric arrow head geometry question. The known parameters are L1, L2, L3, B, and k while the unknowns are t1 and t2. The attempt at a solution involves using the sum of interior angles and the law of cosines, as well as creating a new line L4 to help solve for t1 and t2. The suggestion is to compute the length of L4, the angle t3 between L4 and L2, the length of the top line L5, and the angle t4 between L4 and L3, which can then be used to solve for t1.
  • #1
nucloxylon
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Homework Statement



I have this arrow head geometry question (Please see diagram). I know L1, L2 and L3, angle B and the constant k (notice how the corner angle is equal to k * t2). I don't know t1 and t2. I'm pretty sure I have enough constraints, I'm just having trouble finding the right ones to use. Note that the arrow head is not symmetric so the top edge length is not equal to L3.

Homework Equations


Knowns:
L1
L2
L3
B
k

unknowns:
t1
t2

The Attempt at a Solution



The sum of the interior angles gives me one equation:
360 = t1 + t2+ 180 - t2 + t2 + k*t2 + B
B = 180 - t1 - t2 - kt2

Now I tried using the lengths and forming some equations using the cosine rule. I split the arrow head down the middle (gaining a length Lm), so I have
Lm^2 = L1^2 + L3^2 - 2*L1*L3*cos(t1)

but then I don't have the top edge as I mentioned, so I can't really equate this to anything.

I'm kind of stumped, any help would be appreciated!
 

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  • #2
I would first pretend to know t1 and t2, write down some relations and see if it's possible to solve for t1, t2.
My approach would be to draw a new line L4 that closes the arrow on the left side.
Then you can compute it's length using the law of cosines and the angle t2 (or 180°-t2).
THEN, you've got enough parameters to compute the angle between L4 and L2, let's call it t3.
THEN, you could compute the length of the top line, let's call it L5, using again the law of cosines and the angle B.
THEN, you can compute the angle between L4 and L3, let's call it t4 and you finally have
t1=t4-t3, which "closes the circle". See if you like my approach and if so, try to do the calculations.
 

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Related to Solve Arrow Head Geometry Question with L1, L2 & L3

1. What is Arrow Head Geometry?

Arrow Head Geometry is a mathematical concept that involves the study of the shape and properties of arrow heads. It is commonly used in geometry and trigonometry to calculate the angles and lengths of various arrow head shapes.

2. What are L1, L2, and L3 in relation to Arrow Head Geometry?

In Arrow Head Geometry, L1, L2, and L3 refer to the different sides or lengths of an arrow head shape. L1 is the length of the arrow's shaft, L2 is the length of one of the arrow's arms, and L3 is the length of the other arm.

3. How do you solve an Arrow Head Geometry question using L1, L2, and L3?

To solve an Arrow Head Geometry question using L1, L2, and L3, you can use various formulas and trigonometric functions. For example, to calculate the angle between L1 and L2, you can use the inverse tangent function (tan-1). To find the length of L3, you can use the Pythagorean theorem.

4. What are some real-life applications of Arrow Head Geometry?

Arrow Head Geometry has various real-life applications, such as in architecture, engineering, and even sports. In architecture, it is used to design and construct structures with angled roofs or walls. In engineering, it is used to calculate the angles and lengths of various machine parts. In sports like archery, it is used to optimize the design and performance of arrows.

5. Are there any limitations to using L1, L2, and L3 in Arrow Head Geometry?

While L1, L2, and L3 are useful in solving Arrow Head Geometry questions, they have some limitations. These lengths do not take into account the thickness or width of the arrow head, which can affect the accuracy of calculations. Additionally, they only apply to specific arrow head shapes and may not be applicable to other shapes or objects.

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