Question-44365

Question Number 44365 by ajfour last updated on 27/Sep/18

Commented by ajfour last updated on 27/Sep/18

Find area of △ABC in terms of α, β, and R.

$${Find}\:{area}\:{of}\:\bigtriangleup{ABC}\:{in}\:{terms}\:{of} \\ $$$$\alpha,\:\beta,\:{and}\:{R}. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 28/Sep/18

area =(√(s(s−a)(s−b)(s−c))) (a/(sinα))=(b/(sinβ))=(c/(sin{π−(α+β)}))=2R a=2Rsinα=4Rsin(α/2)cos(α/2) b=2Rsinβ=4Rsin(β/2)cos(β/2) c=2Rsin(α+β)=4Rsin((α+β)/2)cos((α+β)/2) s=((a+b+c)/2)=R{(sinα+sinβ+sin(α+β)} s=R{(2sin((α+β)/2)cos((α−β)/2)+2sin((α+β)/2)cos((α+β)/2))} s=2Rsin((α+β)/2)×2cos(α/2).cos(β/2) s=4Rsin((α+β)/2)cos(α/2)cos(β/2) s−a=4Rcos(α/2)(sin((α+β)/2)cos(β/2)−sin(α/2)) s−b=4Rcos(β/2)(sin((α+β)/2)cos(α/2)−sin(β/2)) s−c=4Rsin((α+β)/2)(cos(α/2)cos(β/2)−cos((α+β)/2)) now putting the value of s,a,b and c in (√(s(s−a)(s−b)(s−c))) it is seen that area of triangle is f(R,α,β)

$${area}\:=\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)}\: \\ $$$$\frac{{a}}{{sin}\alpha}=\frac{{b}}{{sin}\beta}=\frac{{c}}{{sin}\left\{\pi−\left(\alpha+\beta\right)\right\}}=\mathrm{2}{R} \\ $$$${a}=\mathrm{2}{Rsin}\alpha=\mathrm{4}{Rsin}\frac{\alpha}{\mathrm{2}}{cos}\frac{\alpha}{\mathrm{2}} \\ $$$${b}=\mathrm{2}{Rsin}\beta=\mathrm{4}{Rsin}\frac{\beta}{\mathrm{2}}{cos}\frac{\beta}{\mathrm{2}} \\ $$$${c}=\mathrm{2}{Rsin}\left(\alpha+\beta\right)=\mathrm{4}{Rsin}\frac{\alpha+\beta}{\mathrm{2}}{cos}\frac{\alpha+\beta}{\mathrm{2}} \\ $$$${s}=\frac{{a}+{b}+{c}}{\mathrm{2}}={R}\left\{\left({sin}\alpha+{sin}\beta+{sin}\left(\alpha+\beta\right)\right\}\right. \\ $$$${s}={R}\left\{\left(\mathrm{2}{sin}\frac{\alpha+\beta}{\mathrm{2}}{cos}\frac{\alpha−\beta}{\mathrm{2}}+\mathrm{2}{sin}\frac{\alpha+\beta}{\mathrm{2}}{cos}\frac{\alpha+\beta}{\mathrm{2}}\right)\right\} \\ $$$${s}=\mathrm{2}{Rsin}\frac{\alpha+\beta}{\mathrm{2}}×\mathrm{2}{cos}\frac{\alpha}{\mathrm{2}}.{cos}\frac{\beta}{\mathrm{2}} \\ $$$${s}=\mathrm{4}{Rsin}\frac{\alpha+\beta}{\mathrm{2}}{cos}\frac{\alpha}{\mathrm{2}}{cos}\frac{\beta}{\mathrm{2}} \\ $$$${s}−{a}=\mathrm{4}{Rcos}\frac{\alpha}{\mathrm{2}}\left({sin}\frac{\alpha+\beta}{\mathrm{2}}{cos}\frac{\beta}{\mathrm{2}}−{sin}\frac{\alpha}{\mathrm{2}}\right) \\ $$$${s}−{b}=\mathrm{4}{Rcos}\frac{\beta}{\mathrm{2}}\left({sin}\frac{\alpha+\beta}{\mathrm{2}}{cos}\frac{\alpha}{\mathrm{2}}−{sin}\frac{\beta}{\mathrm{2}}\right) \\ $$$${s}−{c}=\mathrm{4}{Rsin}\frac{\alpha+\beta}{\mathrm{2}}\left({cos}\frac{\alpha}{\mathrm{2}}{cos}\frac{\beta}{\mathrm{2}}−{cos}\frac{\alpha+\beta}{\mathrm{2}}\right) \\ $$$${now}\:{putting}\:{the}\:{value}\:{of}\:{s},{a},{b}\:{and}\:{c}\: \\ $$$${in}\:\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)}\:\:{it}\:{is}\:{seen}\:{that}\:{area}\:{of} \\ $$$${triangle}\:{is}\:{f}\left({R},\alpha,\beta\right) \\ $$

Commented by ajfour last updated on 28/Sep/18

$${thanks},\:{anyway}\:{Sir}. \\ $$

Answered by MrW3 last updated on 28/Sep/18

let O=center of circle AC=2×R×sin ((∠AOC)/2)=2R sin β AB=2×R×sin ((∠AOB)/2)=2R sin ∠C=2R sin (α+β) A=Area of ΔABC=((AB×AC×sin α)/2) ⇒A=2R^2 sin α sin β sin (α+β)

$${let}\:{O}={center}\:{of}\:{circle} \\ $$$${AC}=\mathrm{2}×{R}×\mathrm{sin}\:\frac{\angle{AOC}}{\mathrm{2}}=\mathrm{2}{R}\:\mathrm{sin}\:\beta \\ $$$${AB}=\mathrm{2}×{R}×\mathrm{sin}\:\frac{\angle{AOB}}{\mathrm{2}}=\mathrm{2}{R}\:\mathrm{sin}\:\angle{C}=\mathrm{2}{R}\:\mathrm{sin}\:\left(\alpha+\beta\right) \\ $$$${A}={Area}\:{of}\:\Delta{ABC}=\frac{{AB}×{AC}×\mathrm{sin}\:\alpha}{\mathrm{2}} \\ $$$$\Rightarrow{A}=\mathrm{2}{R}^{\mathrm{2}} \mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta\:\mathrm{sin}\:\left(\alpha+\beta\right) \\ $$

Commented by ajfour last updated on 28/Sep/18

Thank you Sir, important result! ⇒ A= 2R^2 (abc)((1/(2R)))^3 ⇒ R= ((abc)/(4A)) .

$${Thank}\:{you}\:{Sir},\:{important}\:{result}! \\ $$$$\Rightarrow\:\:{A}=\:\mathrm{2}{R}^{\mathrm{2}} \left({abc}\right)\left(\frac{\mathrm{1}}{\mathrm{2}{R}}\right)^{\mathrm{3}} \\ $$$$\Rightarrow\:\:\:{R}=\:\frac{{abc}}{\mathrm{4}{A}}\:. \\ $$

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