Question

In triangle ABC, if cotA/2:cotB/2:cotC/2= 1:4:15. Then greatest angle is?

Answer 1

If ABC is a triangle 
A + B + C = π 
=> A/2 + B/2 = π/2 - C/2 
=> tan (A/2 + B/2) = tan (π/2 - C/2) 
=> (tan A/2 + tanB/2) / (1 - tan A/2 tan B/2) = cot C/2 
=> (cot A/2 + cot B/2) / (cot A/2 cot B/2 - 1) = cot C/2 
=> cot A/2 + cot B/2 + cot C/2 = cot A/2 cot B/2 cot C/2 
Of course its only true when ABC are angles of a triangle

Answer 2

Answer:

Step-by-step explanation:

cot(A/2):cot(B/2):cot(c/2) = 1:4:15

√s(s-a)/√(s-b)(s-c) : √s(s-b)/√(s-c)(s-a) : √s(s-c)/√(s-a)(s-b)

                 multiplying all by √(s-a)(s-b)(s-c) and dividing by √s

we get:-

(s-a) : (s-b) : (s-c) = 1:4:15

let common ratio be x

s-a = x; s-b = 4x; s-c = 15x

a = (s-b) + (s-c) = 19x                                    (s-b+s-c = 2s-b-c = a+b+c -b-c = a)

b = (s-a) + (s-c) = 16x

c = (s-b) + (s-a) = 5x

a:b:c = 15x:16x:5x = 15:16:5