A building 150 feet tall casts a 50 foot long shadow. If a person stands at the end of the shadow and looks up to the top of the building, what is the angle of the person's eyes to the top of the building (to the nearest hundredth
Of a degree)? (Assume the person's eyes are 6 feet above ground level.)

Same problem as the previous one, right?

Tan(\Theta) = \frac {(length\,of\,side\,opposite)} {(length\,of\,side\, adjacent)}

The only difference is that you're six feet above the ground. So, draw a diagram. Note that you can subtract 6 feet from the 150 feet to get the length of the side opposite.

Same problem as the previous one, right?

Tan(\Theta) = \frac {(length\,of\,side\,opposite)} {(length\,of\,side\, adjacent)}

The only difference is that you're six feet above the ground. So, draw a diagram. Note that you can subtract 6 feet from the 150 feet to get the length of the side opposite.

OK i drew the diagram

What did you get for Theta?

Tan(\theta)=\frac {150-6}{50} = \frac {144}{50} = 2.88

What did you get for Theta?

Tan(\theta)=\frac {150-6}{50} = \frac {144}{50} = 2.88

I got 2.88 but this is where I'm getting confused

If

tan(\theta) = 2.88

you use the "arctan" function (angle whose tangent is) to get the result.

ArcTan (or arctan) is the same as tan^(-1)

\theta = tan^{-1}(tan(\theta)) = tan^{-1}(2.88) = 70.85\,degrees

I'm not sure how you get values for your arctan function, but it exists on most scientific calculators.

If

tan(\theta) = 2.88

you use the "arctan" function (angle whose tangent is) to get the result.

ArcTan (or arctan) is the same as tan^(-1)

\theta = tan^{-1}(tan(\theta)) = tan^{-1}(2.88) = 70.85\,degrees

I'm not sure how you get values for your arctan function, but it exists on most scientific calculators.

o OK i forgot all about arctan thank you