Trig, find tan A and tan B (no numbers given)

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Homework Help Overview

The problem involves finding the tangent values of angles A and B in a right triangle, where the lengths of the sides are represented by variables rather than specific numbers.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the sides of the triangle and the tangent function, questioning how to derive specific values without numerical inputs. There is mention of the standard tangent definition and the implications of assigning arbitrary values to the sides.

Discussion Status

The discussion is ongoing, with participants exploring the implications of the problem's constraints. Some guidance has been offered regarding the relationships between the tangents of the angles, but no consensus has been reached on specific values or interpretations.

Contextual Notes

There is a lack of numerical information provided in the problem, leading to various interpretations of how to approach the tangent calculations. Participants are considering the implications of this absence on their reasoning.

uperkurk
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Homework Statement



Find tan A, tan B

I have a right angle triangle. Length a = t, length b = k, hypotenuse = p.

The Attempt at a Solution



Usually I would just use the normal rule of [tex]\tan a =\frac{opp}{adj}[/tex] but how can I get an answer here? I can't just say [tex]\tan a =\frac{t}{k}[/tex] because this makes no sense.

The only thing I can think of is say the answer is 1. Or tan a = 45, tan b = 45.
 
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If this is really all you were given, then that is all you can do. The answer cannot possibly be 1 because there is nothing in the problem barring you from making t=2 and k=1, making tan A = 2 (indeed, you could fix it so tan A = any number you want).
 
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Question 6 is the one relating to my original question.
 
Seems to me like all you can do it get the quotients you gave in the first post :)
 
I think your teacher was trying to get you to show that

tanA = 1/tanB

tanA = t/k

tanB = k/t
 
Chestermiller said:
I think your teacher was trying to get you to show that

tanA = 1/tanB

tanA = t/k

tanB = k/t

This. :wink:
 

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