Find Cot A, sec A, csc A for triangle ABC. Let AB= 96.8418, BC=48.4209 and AC=83.865.

Answer 1

AB² = BC² + AC²

So the triangle is the right triangle.

Use SOH CAH TOA to find Angle A

Then Solve for Cot A, Sec A and CSC A using SOH CAH TOA

Answer 2

I confess, I did the problem the long way, then noticed the shortcut.

The long way:

c = AB, b = AC, and a = BC

Then, using the cosine law, we modify the normal version a^2 = b^2 + c^2 - 2bc cos A into

A = arccos((b^2 + c^2 - a^2)/(2bc)) = arccos((83.865^2 + 96.8418^2 - 48.4209^2)/(2*83.865*96.8418)) =

arccos(0.86602540415668) = 30°

cot A = √3

sec A = 2/√3 = 2√3 /3

csc A = 2

The shortcut would be to notice that BC/AB = 1/2 and BC^2 + AC^2 = AB^2 telling us that we have a 30-60-90 triangle. AC should really be 83.8675, though!