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 circum radius of Δ ABC. Given AH.BH.CH = 3 & (AH) 2 + (BH) 2 + (CH) 2 = 7
Then answer the following
(i) Value of 
is
Text Solution
Verified by Experts(i) [using values]; (i) [using values]; (i) [using values]
Ans.
(i)
Sol.
AH = 2R cosA
BH = 2R cos B
CH = 2R cos C
HD = 2R cos B cos C
HE = 2 R cos A cos C
HF = 2R cos A cos B
∴ AH.BH.CH = 3
⇒ Π cos A = 
…...(i) [using values]
Now AH 2 + BH 2 + CH 2 = 7
⇒ 4R 2 ∑ cos 2 A = 7
⇒ ∑ cos 2 A = 
Now we know
cos 2 A + cos 2 B + cos 2 C = 1 – 2 cos A cos B cos C
= 1 – 2 . 
⇒ 4R 3 – 7R–3 = 0
⇒ (R + 1) (2R + 1) (2R –3) = 0
⇒ R = 
Now HD.HE.HF = (2R cos B cos C) (2R cos A cos C)
(2R cos A cos B)
= 8R 3 cos 2 A cos 2 B cos 2 C
= 8R 3 . 
[using (i)] = 
(ii)
Sol.
AH = 2R cosA
BH = 2R cos B
CH = 2R cos C
HD = 2R cos B cos C
HE = 2 R cos A cos C
HF = 2R cos A cos B
∴ AH.BH.CH = 3
⇒ Π cos A = 
…...(i) [using values]
Now AH 2 + BH 2 + CH 2 = 7
⇒ 4R 2 ∑ cos 2 A = 7
⇒ ∑ cos 2 A = 
Now we know
cos 2 A + cos 2 B + cos 2 C = 1 – 2 cos A cos B cos C
= 1 – 2 . 
⇒ 4R 3 – 7R–3 = 0
⇒ (R + 1) (2R + 1) (2R –3) = 0
⇒ R = 
Now HD.HE.HF = (2R cos B cos C) (2R cos A cos C)
(2R cos A cos B)
= 8R 3 cos 2 A cos 2 B cos 2 C
= 8R 3 . 
[using (i)] = 
(iii)
Sol.
AH = 2R cosA
BH = 2R cos B
CH = 2R cos C
HD = 2R cos B cos C
HE = 2 R cos A cos C
HF = 2R cos A cos B
∴ AH.BH.CH = 3
⇒ Π cos A = 
…...(i) [using values]
Now AH 2 + BH 2 + CH 2 = 7
⇒ 4R 2 ∑ cos 2 A = 7
⇒ ∑ cos 2 A = 
Now we know
cos 2 A + cos 2 B + cos 2 C = 1 – 2 cos A cos B cos C
= 1 – 2 . 
⇒ 4R 3 – 7R–3 = 0
⇒ (R + 1) (2R + 1) (2R –3) = 0
⇒ R = 
Now HD.HE.HF = (2R cos B cos C) (2R cos A cos C)
(2R cos A cos B)
= 8R 3 cos 2 A cos 2 B cos 2 C
= 8R 3 . 
[using (i)] = 
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