Why do you use cot in this eqation, my first thought was to use tan.

My understanding is cotθ = a/o and tanθ = o/a, would I get the same results if I used tanθ/2 = b/2 /d is there a reason my online teacher defaults to cot for this one?

CPhill helped me on a previous question wasn't sure how to thank him. So early thanks to anyone who helps me <3

Guest Sep 16, 2019

#1**+2 **

We don't need the cotangent.....notice that

tan (θ/2) = opp/adj = (b/2) / d = 1 cm / d

The only difficult thing here is to convert θ = 1° 23 ' 12 " to decimal degrees.....so we have

[1 + (23/60) + (12/3600)]° ≈ 1.387°

So θ/2 = (1.387° / 2) = (.6935°)

So...we have

tan (.6935°) = 1 / d rearrange as

d = 1 / tan (.6935°) ≈ 82.6 cm

CPhill Sep 16, 2019