Published by:
CGP EDU Academic Team
Published on: August 14, 2026
An acute angled triangle ABC is inscribed in a circle of radius R. The altitudes from the vertices A, B, C meet the circle at D, E, F respectively. Let s', r' and r' 1 be the semiperimeter, inradius and exradius opposite to D of triangle DEF. Then
(i) 
=
A
Σ sin 2A 1 – Σ cos2A 1 + cos2A – cos2B – cos2C (b – cos2A + cos2B + cos2C
B
Σ sinA –1 – Σ cos2A
C
Σ cosA 1 + Σ sinA 1 + sin2A – sin2B – sin 2C
D
Σ cos2A (ii) 
= –1 + Σ cos2A (iii) 
= 2sinA – sinB – sinC
= –1 + Σ cos2A (iii) = 2sinA – sinB – sinC
Text Solution
Verified by ExpertsThe correct answer is:
D
Ans.
(i)
Sol. D = π –2A, E = π –2B, F = π –2C
EF = 2R sin D = 2R sin 2A
∴ 
= Σ sin 2A
(ii)
Sol. r ′ = 4R sin 
sin 
sin 
= 4 cos A cos B cos C = –1 – Σ cos 2A
(iii)
Sol. 
= 4R sin 
cos 
cos 
= 4 cos A sin B sin C
= 1 + cos 2A – cos 2B – cos 2C
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