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Question Number 123697 by ajfour last updated on 27/Nov/20

Commented by ajfour last updated on 27/Nov/20

Find radius of quarter circle in terms of sides a, b, c of △ABC.

Findradiusofquartercircleintermsofsidesa,b,cof△ABC.

Answered by mr W last updated on 27/Nov/20

Commented by mr W last updated on 27/Nov/20

B(0,0) A(c cos β,c sin β) C(a,0) x_D =(R/(tan β)) eqn. of AC: (y/(x−a))=((c sin β)/(c cos β−a)) c sin β x+(a−c cos β)y−ac sin β=0 R^2 =(([c sin β (R/(tan β))−ac sin β]^2 )/(c^2 sin^2 β+(a−c cos β)^2 )) R^2 =(((R cos β−a sin β)^2 c^2 )/(a^2 +c^2 −2ac cos β)) R^2 =(((R^2 cos^2 β+a^2 sin^2 β−aR sin 2β)c^2 )/b^2 ) (c^2 cos^2 β−b^2 )R^2 −ac^2 sin 2β R+a^2 c^2 sin^2 β=0 ⇒R=((ac sin β)/(c cos β+b)) =((2a^2 c sin β)/(a^2 +c^2 −b^2 +2ab)) R=((4aΔ)/(a^2 +c^2 −b^2 +2ab)) with Δ=(√(s(s−a)(s−b)(s−c))) example: c=a, b=(√2)a Δ=(a^2 /2) ⇒R=((4a×(a^2 /2))/(2(√2)a))=(a/( (√2))) ✓

B(0,0)A(ccosβ,csinβ)C(a,0)xD=Rtanβeqn.ofAC:yx−a=csinβccosβ−acsinβx+(a−ccosβ)y−acsinβ=0R2=[csinβRtanβ−acsinβ]2c2sin2β+(a−ccosβ)2R2=(Rcosβ−asinβ)2c2a2+c2−2accosβR2=(R2cos2β+a2sin2β−aRsin2β)c2b2(c2cos2β−b2)R2−ac2sin2βR+a2c2sin2β=0⇒R=acsinβccosβ+b=2a2csinβa2+c2−b2+2abR=4aΔa2+c2−b2+2abwithΔ=s(s−a)(s−b)(s−c)example:c=a,b=2aΔ=a22⇒R=4a×a2222a=a2✓

Commented by ajfour last updated on 27/Nov/20

let A(h,k) ; D(p,0) (R/p)=(k/h) ..(i) eq. of AC : y=((k/(a−h)))(a−x) ((k(a−p))/( (√((a−h)^2 +k^2 ))))=R ...(ii) ⇒ k(a−((hR)/k))=R(√((a−h)^2 +k^2 )) R=((ak)/( h+(√((a−h)^2 +k^2 )))) R=((acsin β)/(ccos β+b)) .

letA(h,k);D(p,0)Rp=kh..(i)eq.ofAC:y=(ka−h)(a−x)k(a−p)(a−h)2+k2=R...(ii)⇒k(a−hRk)=R(a−h)2+k2R=akh+(a−h)2+k2R=acsinβccosβ+b.

Commented by ajfour last updated on 27/Nov/20

Thanks Sir, I managed it too.

ThanksSir,Imanagedittoo.

Commented by mr W last updated on 27/Nov/20

very nice!

verynice!

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