Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

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.

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

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)) .

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com