Recent content by jaycob1997

Another approach to this problem is to use the formula for changing the base, which is: Log(base x)N=(Log(base 10)N)/(Log(base 10)x). This can be very useful in any future quizzes or exams that you may have. This is also applicable for Ln (which is Log(base e)), e=2.7182... Applying...

I did further reading on the subject matter. This is where I got lost, the problem goes: In a right spherical triangle (C=90 degrees), A=69 degrees and 50.8minutes, c=72 degrees and 15.4minutes, find B, a, b. Using Napier's rules, I get a=63 degrees and 23.8 minutes b= 47 degrees7.0...

Thanks for the link. Do you know other sites that I can go to regarding the same subject matter (Spherical Trigonometry), or if you know of any sites that have problems to solve so I can apply the concepts that I just learned. By the way, I'm self studying for my licensure examination, and a...

Let me just add some other information: I'm using Napier's rules to solve for spherical angles and spherical sides for right spherical triangles. Now my problem is, when will I use the acute angles and not the supplement angles(Obtuse). Or are there any rule/s to follow on which angle to use.

I'm trying to review spherical trinometry on my own and I'm stuck where it says the supplement of the solved dihedral angle can be implemented. I'm using "PLANE AND SPHERICAL TRIGONOMETRY" by Paul Rider. Can anyone explain the concept behind that statement, because the examples that I have are...