+0  
 
0
218
2
avatar

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

 Sep 16, 2019
 #1
avatar+111396 
+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

 

cool cool cool

 Sep 16, 2019
 #2
avatar
+1

Thanks I feel like I better understand the relation now

Guest Sep 16, 2019

11 Online Users