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Question Number 201146 by ajfour last updated on 30/Nov/23

Commented by ajfour last updated on 30/Nov/23

$H o w f a r i s J f r o m c e n t e r o f c i r c l e ?$
Commented by mr W last updated on 01/Dec/23

Commented by ajfour last updated on 02/Dec/23

$T h a n k s f o r t h e c o i n i n g, w o n t y o u$ $a t t e m p t t h e q u e s t i o n s i r ?$
Commented by mr W last updated on 02/Dec/23

$t o g e t a g e n e r a l f o r m u l a i s v e r y$ $h a r d, i t h i n k .$
	Answered by mr W last updated on 04/Dec/23

	Commented by mr W last updated on 04/Dec/23

	$r = r a d i u s o f i n c i r c l e$ $r = \frac{2 Δ}{a + b + c}$ $v + w = a$ $w + u = b$ $u + v = c$ $\Rightarrow u + v + w = \frac{a + b + c}{2}$ $\Rightarrow u = \frac{- a + b + c}{2}, v = \frac{a - b + c}{2}, w = \frac{a + b - c}{2}$ $v \sin B = (c - v \cos B) \tan α$ $\Rightarrow \tan α = \frac{v \sin B}{c - v \cos B}$ $s i m i l a r l y$ $\Rightarrow \tan β = \frac{w \sin C}{a - w \cos C}$ $ϕ = π - B - α$ $D J \sin ϕ = (v - D J \cos ϕ) \tan β$ $\Rightarrow D J = \frac{v \tan β}{\sin ϕ + \cos ϕ \tan β}$ $\Rightarrow D J = \frac{v \tan β}{\sin (α + B) - \cos (α + B) \tan β}$ ${J I}^{2} = {D J}^{2} + r^{2} - 2 r D J \sin ϕ$ $J I = \sqrt{{D J}^{2} + r^{2} - 2 r D J \sin (α + B)}$
	Commented by ajfour last updated on 04/Dec/23

	$T h a n k s s i r, b e t t e r, b u t c a n b e$ $i m p r o v e d s t i l l, i s h a l l t r y .$
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