AGNuke
Gold Member
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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.
[tex]\frac{\prod \cos A}{\sum \cos^{2}A}[/tex]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.
Given AH.BH.CH = 3 and (AH)2 + (BH)2 + (CH)2 = 7, answer the following
Q1.
[tex]\frac{\prod \cos A}{\sum \cos^{2}A}[/tex]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.