- #1

thunderjolt

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## Homework Statement

I've solved a problem in my engineering homework to a point where I have the following data:

ψ = arctan(3) = 71.57 (degrees) , inverse tangent of 3

δ = 30 degrees

solve for gamma and chi (χ):

tan(2gamma) = tan(2ψ)cos(δ)

sin(2χ) = sin(2ψ)sin(δ)

## Homework Equations

tan(2gamma) = tan(2ψ)cos(δ)

sin(2χ) = sin(2ψ)sin(δ)

## The Attempt at a Solution

My problem rests with the gamma solution. When I draw it (using other data in the problem), it is an angle larger than 70 degrees at least. When I solve for gamma with a calculator I get gamma = -16.5°.

my process:

tan(2ψ) = -.75

(-.75)cos(30) = -0.6495

arctan(-0.6495) = -33°

gamma = -33/2 = -16.5°

I know that there are properties of tangent that you need to take into consideration, but I just cannot remember them, and I am sure that is why I get this negative angle.

Or, did I do this correctly, and I just need to redraw my problem?

EDIT: the answer in the back of the book is gamma = 73.5 degrees

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