Trigonometric functions problem

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Homework Help Overview

The problem involves finding the value of x in a triangle defined by specific dimensions and angles, utilizing trigonometric functions, particularly tangent. The context includes a triangle with a base composed of two segments and an angle related to the opposite side.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply trigonometric identities and equations to express relationships between the sides and angles of the triangle. Some participants question the accuracy of the algebraic manipulation, particularly regarding the loss of a variable in the equation.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's algebraic steps. There is a recognition of a potential error in the calculations, which has led to a clarification that may help resolve the confusion. However, no consensus has been reached on the final approach to the problem.

Contextual Notes

Participants are navigating through algebraic complexities and trigonometric identities, with some expressing confusion about the manipulation of the equations. The original poster indicates a desire to use a specific mathematical technique, but feels it may complicate the situation further.

luludatis
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Homework Statement


Find x

base of triangle: 2 pieces, 8+12
opposite end: x
angle: [tex]\Theta[/tex] ; angle formed by triangle of 8 units and x : 2[tex]\Theta[/tex]


Homework Equations


tan[tex]\Theta[/tex] = opp/adj
tan 2[tex]\Theta[/tex] = 2tan/1-tan2

The Attempt at a Solution


tan[tex]\Theta[/tex]= x/20
tan 2[tex]\Theta[/tex] = x/8

x/8=tan2[tex]\Theta[/tex]= 2(x/20) / 1-(x/20)2

(2x/20) / ((400-x2)/400) = 40x/400-x2

x/8=40x/400-x2

320x=400x-x3
x3-80=0


and now I'm stuck. I want to use difference of cubes, but i feel it is getting too complicated...
 

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welcome to pf!

hi luludatis! welcome to pf! :smile:
luludatis said:
320x=400x-x3
x3-80=0

oooh :cry: …

you lost an x ! :blushing:​
 
...i'm confused :P
 
x3 - 80x = 0 :wink:
 
oooooooooooooooh! great! that fixes everything! thank you!
 

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