• Jan 27th 2013, 03:56 PM
hi, im new to this website, I really love math i've just started actually learning it a few years ago.

but anyways im getting ready to take readiness test for calc i need help with this question
its been a while since i did pre calc i just got off a lazy k winter break and need to get back into shape
indepth explanation of the answer would help alot THANK YOU IN ADVANCE.

Let R be the radius of the circle circumscribed in the triangle of sides, 1968, 1968,1968, and let r denote the radius of the circle inscribed in this triangle
Then the ratio is R/r is
a. 2.
b. tan π/6
c. cos π/3
d. 1968
e.√3/2
• Jan 27th 2013, 04:58 PM
Plato
Quote:

Let R be the radius of the circle circumscribed in the triangle of sides, 1968, 1968,1968, and let r denote the radius of the circle inscribed in this triangle
Then the ratio is R/r is
a. 2.
b. tan π/6
c. cos π/3
d. 1968
e.√3/2

You have an equilateral triangle.

Look at that webpage. You can see the answer.
• Jan 27th 2013, 05:42 PM
So, R/r= √3/3 *(1968) divided by √3/6 *(1968)
• Jan 27th 2013, 05:53 PM
Plato
Quote:

So, R/r= √3/3 *(1968) divided by √3/6 *(1968)