Find the Circumradius for a Triangle

[AGNuke]

Let ABC is an acute angled triangle with orthocentre H. D, E, F are feet of perpendicular from A, B, C on opposite sides. Let R is circumradius of ΔABC.

Given AH.BH.CH = 3 and (AH)² + (BH)² + (CH)² = 7, answer the following
Q1.
$$\frac{\prod \cos A}{\sum \cos^{2}A}$$Q2. What is the value of R?

ANS 1. From properties of triangle, the distance of Orthocentre from a point A is given by AH = 2R.cosA. Using the values of cosines and from information in the question, I solved the first question to get the answer 3/14R.

Now I have no clue on how to approach Q2. I can't seem to find any relation between the value of R and information given to me. BTW, from what I know, the answer mentioned is 3/2.

 

[Simon Bridge]

What does the answer to Q1 tell you?
Why is it important that the triangle is acute angled?

 

[AGNuke]

Acute angled triangle means all the cosines are greater than zero?