How many triangles can be formed from non−adjacent vertices of a regular polygon that it angle is 177^( °) ?


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Question Number 180773 by Acem last updated on 17/Nov/22

$H o w m a n y t r i a n g l e s c a n b e f o r m e d f r o m$ $n o n - a d j a c e n t v e r t i c e s o f a r e g u l a r p o l y g o n$ $t h a t i t a n g l e i s 177^{°} ?$
	Answered by mr W last updated on 17/Nov/22

	$n = \frac{360}{180 - 177} = 120 v e r t i c e s$ $w e c a n f o r m$ $\frac{1}{3} \times n \times (C_{2}^{n - 3} - (n - 4))$ $= \frac{n (n - 4) (n - 5)}{6}$ $= 266 800 t r i a n g l e s$
	Commented by Acem last updated on 17/Nov/22

	$M o r n i n g S i r!, t h e n u m b e r i s r a t h e r c l o s e t o$ $t h e c o r r e c t a n s w e r, i f w e a p p l y t h e f o r m u l a a b o v e$ $t o a h e x a g o n o r o t h e r w i s e w e {w o u l d n}^{'} t g e t t h e$ $r e a l a n s w e r .$
	Commented by mr W last updated on 17/Nov/22

	$i s e e m y m i s t a k e n o w . i^{'} l l f i x i t .$ $t h e v e r t r i c e s o f t r i a n g l e s h o u l d n o t$ $b e a d j a c e n t v e r t r i c e s o f t h e p o l y g o n .$ $i {d i d n}^{'} t f o l l o w t h i s c o r r e c t l y .$
	Commented by Acem last updated on 17/Nov/22

	$Y e s S i r, y o u w i l l d o i t! t h a n k y o u$ $N o t e : T h e n u m b e r \in] 265 349, 266 801 [$
	Commented by mr W last updated on 17/Nov/22

	$i g o t 266800 . p l e a s e c h e c k m y f i x e d$ $s o l u t i o n .$
	Commented by Acem last updated on 17/Nov/22
	$Shakehands! Whatever Regular Polygon { ((C_3 ^( n) − n(n−3))),(( ((n(n−4)(n−5))/6))) :} ∗ If we want to count only the triangles that are formed from adjacent vertices, apply n(n−3)$
	$S h a k e h a n d s!$ $W h a t e v e r R e g u l a r P o l y g o n {\begin{cases} C_{3}^{n} - n (n - 3) \\ \frac{n (n - 4) (n - 5)}{6} \end{cases}$ $* I f w e w a n t t o c o u n t o n l y t h e t r i a n g l e s t h a t a r e$ $f o r m e d f r o m a d j a c e n t v e r t i c e s, a p p l y n (n - 3)$
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