In-ABC-cos-A-cos-B-cos-C-3-2-prove-that-trianle-is-equilateral-

Question Number 54502 by math1967 last updated on 05/Feb/19

I n △ A B C \cos A + \cos B + \cos C = \frac{3}{2}

p r o v e t h a t t r i a n l e i s e q u i l a t e r a l

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Feb/19

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Feb/19

p o i n t A, B, C l i e s o n c o s x c u r v e

p o i n t A c o o r d i n a t e i s (A, c o s A)

p o i n t B c o o r d i n a t e (B, c o s B)

p o i n t C c o o r d i n a t e (C, c o s C)

△ A B C c e n t r o i d i s G

c o o r d i n a t e o f G i s (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})

s o G i s (\frac{A + B + C}{3}, \frac{c o s A + c o s B + c o s C}{3})

a s p e r d i a g r a m p o i n t Q (\frac{A + B + C}{3}, 0)

p o i n t p {\frac{A + B + C}{3}, c o s (\frac{A + B + C}{3})}

f r o m d i a g r a m

Q P > Q G

c o s (\frac{A + B + C}{3}) > \frac{c o s A + c o s B + c o s C}{3}

3 c o s (\frac{π}{3}) > c o s A + c o s B + c o s C

c o s A + c o s B + c o s C < \frac{3}{2}

n o w m a x i m u m v a l u e o f c o s A + c o s B + c o s C

w h e n A = B = C \dots s o A = B = C = \frac{π}{3}

s o c o s \frac{π}{3} + c o s \frac{π}{3} + c o s \frac{π}{3}

= \frac{1}{2} + \frac{1}{2} + \frac{1}{2}

= \frac{3}{2} t h a t m e a n s e q u i l a t e r a l t r i a n g l e \dots

[i n o t h e r c a s e s c o s A + c o s B + c o s C < \frac{3}{2}]

Commented by math1967 last updated on 05/Feb/19

T h a n k y o u s i r

Commented by math1967 last updated on 05/Feb/19

o t h e r m e t h o d

\cos (A + B) + \cos (B + C) + \cos (C + A) = - \frac{3}{2}

\Rightarrow {(\sin A - \sin B)}^{2} + {(\sin B - \sin C)}^{2} + {(\sin C - \sin A)}^{2}

{(\cos A + \cos B)}^{2} + {(\cos B + \cos C)}^{2} + {(\cos C + \cos A)}^{2} = 0

∴ {(\sin A - \sin B)}^{2} = 0 \Rightarrow A = B

s i m i l a r l y B = C ∴ △ A B C e q u i l a t e r a l

i f \cos A + \cos B = 0 t h e n A = π - B

n o t p o s s i b l e f o r t r i a n g l e

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Feb/19

b a h d a r u n \dots