Two Simultaneous Eqns with Trig, Statcs Problem

[birdec]

Homework Statement

I'm doing a statics problem and am following how the solutions manual does it but they skipped a few steps and I'm lost as a result.

I have two equations and two unknowns: F and θ (degrees)

Homework Equations

Equation 1: F*cos(25+θ) = -54.684
Equation 2: F*sin(25+θ) = 69.131

The Attempt at a Solution

Solve for F: F=-54.684/cos(25+θ)

Plug F into other equation and combine sin and cos to get tangent(25+θ):

-54.684*tan(25+θ)=69.131
tan(25+θ)=(-1.26)
25+θ=arctan(-1.26)
25+θ=-51.65
θ=-76.65

The solutions manual simply says "Solving Equations (1) and (2) yields 25+θ=128.35; θ=103 degrees; F = 88.1 lb"

The only thing I can think of is adding 180 degrees to the θ I got which gives me the 103 degrees they got. Any ideas?

 

[Mark44]

Homework Statement

I'm doing a statics problem and am following how the solutions manual does it but they skipped a few steps and I'm lost as a result.

I have two equations and two unknowns: F and θ (degrees)

Homework Equations

Equation 1: F*cos(25+θ) = -54.684
Equation 2: F*sin(25+θ) = 69.131

The Attempt at a Solution

Solve for F: F=-54.684/cos(25+θ)

Plug F into other equation and combine sin and cos to get tangent(25+θ):

-54.684*tan(25+θ)=69.131
tan(25+θ)=(-1.26)
25+θ=arctan(-1.26)
25+θ=-51.65
θ=-76.65

The solutions manual simply says "Solving Equations (1) and (2) yields 25+θ=128.35; θ=103 degrees; F = 88.1 lb"

The only thing I can think of is adding 180 degrees to the θ I got which gives me the 103 degrees they got. Any ideas?

Sounds good to me. The tangent function is periodic with period π (or 180°), so adding 180° to what you got will produce an angle in the second quadrant. Presumably any angle you get should be between 0° and 180°.

BTW, you could have shortened your work by dividing the 2nd equation by the first. That would have eliminated F in one step and gotten you directly to tan(25 + θ) on one side of the equation.