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Question Number 22116 by ajfour last updated on 11/Oct/17

Commented by ajfour last updated on 11/Oct/17

to prove : (b+c)^2 ≥ a^2 +4h_a ^2 . see Q.22079

toprove:(b+c)2⩾a2+4ha2.seeQ.22079

Answered by ajfour last updated on 11/Oct/17

a^2 =b^2 +c^2 −2bc cos (θ+φ) =(b+c)^2 −2bc[1+cos (θ+φ)] =(b+c)^2 −2bc[1+2cos θcos φ −cos θcos φ −sin θsin φ] =(b+c)^2 −2bc[1+2×(h_a /c).(h_a /b)−cos (θ−φ)] =(b+c)^2 −4h_a ^2 −2bc[1−cos (θ−φ)] so (b+c)^2 −a^2 −4h_a ^2 =4bcsin^2 (((θ−φ)/2))≥0 .

a2=b2+c2−2bccos(θ+ϕ)=(b+c)2−2bc[1+cos(θ+ϕ)]=(b+c)2−2bc[1+2cosθcosϕ−cosθcosϕ−sinθsinϕ]=(b+c)2−2bc[1+2×hac.hab−cos(θ−ϕ)]=(b+c)2−4ha2−2bc[1−cos(θ−ϕ)]so(b+c)2−a2−4ha2=4bcsin2(θ−ϕ2)⩾0.

Commented by Tinkutara last updated on 11/Oct/17

Thank you very much Sir!

ThankyouverymuchSir!

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