Aditi Chauhan
Last Activity: 10 Years ago
Hello Student,
Please find the answer to your question
In ∆ ACD, cos C = b/a/2 = 2b/a . . . . . . . . . . . . . . . . . . . . . . . . (1)
In ∆ ABC, cos C = a2 + b2 + c2/2ab . . . . . . . . . . . . . . . . . . . . . . . . (2)
From (1) and (2),
2b/a = a2 + b2 - c2/2ab ⇒ b2 = 1/3 (a2 – c2) . . . . . . . . . . . . . . (3)
Also cos A = b2 + c2 – a2/2bc
∴ cos A cos C = b2 + c2 – a2/2bc x 2b/a = b2 + c2 – a2/ac
=

= 2(c
2 – a
2)/3ac
Thanks
Aditi Chauhan
askIITians Faculty