in AB^Δ C : cos (A)+cos(B)+cos(C)=(7/4) (R/r) =?


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Question Number 170856 by mnjuly1970 last updated on 01/Jun/22

$i n A \overset{Δ}{B C} : c o s (A) + c o s (B) + c o s (C) = \frac{7}{4}$ $\frac{R}{r} = ?$
Commented by mr W last updated on 01/Jun/22

$\cos A + \cos B + \cos C = \frac{7}{4} n o t p o s s i b l e,$ $s i n c e \cos A + \cos B + \cos C ⩽ \frac{3}{2} .$
Commented by mnjuly1970 last updated on 02/Jun/22

$y e s y o u a r e r i g h t s i r . . .$ $i a p p o l o g i z e . . . i t h i n k . . .$ $s i n (A) + s i n (B) + s i n (B) \overset{?}{=} 1 + \frac{r}{R}$ $\overset{R ⩾ 2 r}{⩽} 1 + \frac{1}{2} = \frac{3}{2}$
	Answered by mr W last updated on 01/Jun/22

	$a s s u m e \cos A + \cos B + \cos C = \frac{5}{4} .$ $a = 2 R \sin A, b = 2 R \sin B, c = 2 R \sin C$ $Δ = \frac{a b \sin C}{2}$ $r = \frac{2 Δ}{a + b + c}$ $\frac{R}{r} = \frac{R (a + b + c)}{2 Δ}$ $\frac{R}{r} = \frac{R (a + b + c)}{a b \sin C}$ $\frac{R}{r} = \frac{2 R^{2} (\sin A + \sin B + \sin C)}{4 R^{2} \sin A \sin B \sin C}$ $\frac{R}{r} = \frac{\sin A + \sin B + \sin C}{2 \sin A \sin B \sin C}$ $\frac{R}{r} = \frac{4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}}{16 \sin \frac{A}{2} \cos \frac{C}{2} \sin \frac{B}{2} \cos \frac{B}{2} \sin \frac{C}{2} \cos \frac{C}{2}}$ $\frac{R}{r} = \frac{1}{4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}}$ $\frac{R}{r} = \frac{1}{\cos A + \cos B + \cos C - 1}$ $\frac{R}{r} = \frac{1}{\frac{5}{4} - 1} = 4 ✓$
	Commented by MJS_new last updated on 01/Jun/22

	$I think {there}^{'} s no such triangle$
	Commented by mr W last updated on 01/Jun/22

	$y o u a r e r i g h t s i r!$ $\cos A + \cos B + \cos C m u s t b e ⩽ \frac{3}{2} .$
	Commented by mnjuly1970 last updated on 02/Jun/22

	$t h a n k s a l o t . . .$
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