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Question Number 75058 by behi83417@gmail.com last updated on 07/Dec/19
in triangle: ABC: a=(√(2 )),b−c=(((√2)+1)/2),B^� −C^� =(𝛑/2) find: h_a , S_(ABC ) ,d_(a ) , R ,A^� .
$$\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{triangle}}:\:\:\boldsymbol{\mathrm{ABC}}: \\ $$$$\boldsymbol{\mathrm{a}}=\sqrt{\mathrm{2}\:},\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{c}}=\frac{\sqrt{\mathrm{2}}+\mathrm{1}}{\mathrm{2}},\overset{} {\boldsymbol{\mathrm{B}}}−\overset{} {\boldsymbol{\mathrm{C}}}=\frac{\boldsymbol{\pi}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{find}}:\:\:\boldsymbol{\mathrm{h}}_{\boldsymbol{\mathrm{a}}} ,\:\:\boldsymbol{\mathrm{S}}_{\boldsymbol{\mathrm{ABC}}\:\:} ,\boldsymbol{\mathrm{d}}_{\boldsymbol{\mathrm{a}}\:\:\:} ,\:\boldsymbol{\mathrm{R}}\:\:\:\:,\overset{} {\boldsymbol{\mathrm{A}}}. \\ $$
Commented by mr W last updated on 07/Dec/19
b−c=(1/2) is not possible sir! it must be 1<b−c<(√2) !
$${b}−{c}=\frac{\mathrm{1}}{\mathrm{2}}\:{is}\:{not}\:{possible}\:{sir}! \\ $$$${it}\:{must}\:{be}\:\mathrm{1}<{b}−{c}<\sqrt{\mathrm{2}}\:! \\ $$
Commented by behi83417@gmail.com last updated on 07/Dec/19
thank you very much dear master. now it is fixed.
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{dear}\:\mathrm{master}. \\ $$$$\mathrm{now}\:\mathrm{it}\:\mathrm{is}\:\mathrm{fixed}. \\ $$
Answered by mr W last updated on 07/Dec/19
let ∠C=θ ⇒∠B=(π/2)+θ ⇒∠A=π−θ−((π/2)+θ)=(π/2)−2θ (b/(sin ((π/2)+θ)))=(c/(sin θ))=((√2)/(sin ((π/2)−2θ))) ⇒b=(((√2) cos θ)/(cos 2θ)) ⇒c=(((√2) sin θ)/(cos 2θ)) b−c=(((√2) (cos θ−sin θ))/(cos 2θ))=δ ((2 (1−sin 2θ))/(cos^2 2θ))=δ^2 (1/(1+sin 2θ))=(δ^2 /2) sin 2θ=(2/δ^2 )−1 ⇒1<δ<(√2) ⇒𝛉=(1/2)sin^(−1) ((2/𝛅^2 )−1) with δ=((1+(√2))/2) we get θ=(1/2)sin^(−1) (23−16(√2))≈10.9475° the rest is clear.
$${let}\:\angle{C}=\theta \\ $$$$\Rightarrow\angle{B}=\frac{\pi}{\mathrm{2}}+\theta \\ $$$$\Rightarrow\angle{A}=\pi−\theta−\left(\frac{\pi}{\mathrm{2}}+\theta\right)=\frac{\pi}{\mathrm{2}}−\mathrm{2}\theta \\ $$$$\frac{{b}}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}+\theta\right)}=\frac{{c}}{\mathrm{sin}\:\theta}=\frac{\sqrt{\mathrm{2}}}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\mathrm{2}\theta\right)} \\ $$$$\Rightarrow{b}=\frac{\sqrt{\mathrm{2}}\:\mathrm{cos}\:\theta}{\mathrm{cos}\:\mathrm{2}\theta} \\ $$$$\Rightarrow{c}=\frac{\sqrt{\mathrm{2}}\:\mathrm{sin}\:\theta}{\mathrm{cos}\:\mathrm{2}\theta} \\ $$$${b}−{c}=\frac{\sqrt{\mathrm{2}}\:\left(\mathrm{cos}\:\theta−\mathrm{sin}\:\theta\right)}{\mathrm{cos}\:\mathrm{2}\theta}=\delta \\ $$$$\frac{\mathrm{2}\:\left(\mathrm{1}−\mathrm{sin}\:\mathrm{2}\theta\right)}{\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}\theta}=\delta^{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+\mathrm{sin}\:\mathrm{2}\theta}=\frac{\delta^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{sin}\:\mathrm{2}\theta=\frac{\mathrm{2}}{\delta^{\mathrm{2}} }−\mathrm{1}\:\:\:\:\Rightarrow\mathrm{1}<\delta<\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\boldsymbol{\theta}=\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{sin}}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\boldsymbol{\delta}^{\mathrm{2}} }−\mathrm{1}\right) \\ $$$${with}\:\delta=\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}}\:{we}\:{get} \\ $$$$\theta=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{23}−\mathrm{16}\sqrt{\mathrm{2}}\right)\approx\mathrm{10}.\mathrm{9475}° \\ $$$${the}\:{rest}\:{is}\:{clear}. \\ $$

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