Question Number 170010 by mr W last updated on 13/May/22
Commented by mr W last updated on 13/May/22
Answered by mr W last updated on 15/May/22
Commented by mr W last updated on 15/May/22
![we can see that ΔQRP∼ΔABC ⇒(p/c)=(q/a)=(r/b)=k, say ((BP)/(sin (B+θ)))=((RP)/(sin B)) ⇒BP=((ka sin (B+θ))/(sin B)) ((PC)/(sin θ))=((PQ)/(sin C)) ⇒PC=((kb sin θ)/(sin C)) a=((ka sin (B+θ))/(sin B))+((kb sin θ)/(sin C)) ⇒a×((sin B−k sin (B+θ))/(sin B))=((kb sin θ)/(sin C)) (i) similarly ⇒b×((sin C−k sin (C+θ))/(sin C))=((kc sin θ)/(sin A)) (ii) ⇒c×((sin A−k sin (A+θ))/(sin A))=((ka sin θ)/(sin B)) (iii) (i)×(ii)×(iii): [sin A−k sin (A+θ)][sin B−k sin (B+θ)][sin C−k sin (C+θ)]=sin^3 θ k^3 [sin^3 θ+sin (A+θ) sin (B+θ) sin (C+θ)]k^3 −[sin A sin (B+θ) sin (C+θ)+sin B sin (C+θ) sin (A+θ)+sin C sin (A+θ) sin (B+θ)]k^2 +[sin A sin B sin (C+θ)+sin (A+θ) sin B sin C+sin A sin B sin (C+θ)]k −sin A sin B sin C=0 sin^3 θ+sin (A+θ) sin (B+θ) sin (C+θ) =sin^3 θ+(sin A cos θ+cos A sin θ)(sin B cos θ+cos B sin θ)(sin C cos θ+cos C sin θ) =sin^3 θ+[sin A sin B−cos C sin^2 θ+sin C sin θ cos θ](sin C cos θ+cos C sin θ) =sin^3 θ+sin A sin B sin C cos θ− sin^3 θ+(cos A cos B cos C+1)sin θ =sin A sin B sin C cos θ+(cos A cos B cos C+1) sin θ sin A sin (B+θ) sin (C+θ) =sin A (sin B cos θ+cos B sin θ)(sin C cos θ+cos C sin θ) =sin A (sin B sin C cos^2 θ+sin A sin θ cos θ+cos B cos C sin^2 θ) =sin A sin B sin C cos^2 θ+sin^2 A sin θ cos θ+(1/4)(sin 2A+sin 2B− sin 2C) sin^2 θ Σ=3 sin A sin B sin C cos^2 θ+(sin^2 A+sin^2 B+sin^2 C)sin θ cos θ +(3/4)(sin 2A+sin 2B+ sin 2C) sin^2 θ−(1/2)(sin 2A+sin 2B+sin 2C) sin^2 θ =sin A sin B sin C (cos 2θ+2)+(cos A cos B cos C+1) sin 2θ sin A sin B sin (C+θ) sin A sin B (sin C cos θ+cos C sin θ) sin A sin B sin C cos θ+(1/2)(sin^2 A+sin^2 B+sin^2 C)sin θ−sin^2 C sin θ Σ=3 sin A sin B sin C cos θ+(cos A cos B cos C+1) sin θ {sin A sin B sin C cos θ+(cos A cos B cos C+1) sin θ}k^3 −{sin A sin B sin C (cos 2θ+2)+(cos A cos B cos C+1) sin 2θ}k^2 +{3 sin A sin B sin C cos θ+(cos A cos B cos C+1) sin θ}k −sin A sin B sin C=0 or let λ=((cos A cos B cos C+1)/(sin A sin B sin C))=((a^2 +b^2 +c^2 )/(4Δ_(ABC) )) (cos 𝛉+𝛌 sin 𝛉)k^3 −(2+cos 2𝛉+𝛌 sin 2𝛉)k^2 +(3 cos 𝛉+𝛌 sin 𝛉)k−1=0 ⇒ determinant (((k=(1/(cos 𝛉+𝛌 sin 𝛉))))) (wow! i didn′t expect that the root of this complicated cubic equation is so brief!) Δ_(PQR) =k^2 Δ_(ABC) 𝚫_(PQR) =(𝚫_(ABC) /((cos 𝛉+𝛌 sin 𝛉)^2 )) special case: θ=(π/2) ⇒k=(1/λ)=((sin A sin B sin C)/(cos A cos B cos C+1)) this is the same as we get in Q169815.](https://www.tinkutara.com/question/Q170061.png)
Commented by Tawa11 last updated on 08/Oct/22
Question-170010