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Ratio-of-green-blue-and-Red-triangles-Evaluate-these-Area-for-25-and-length-of-aquart-1-

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Question Number 203171 by a.lgnaoui last updated on 11/Jan/24
Ratio of green blue and Red triangles? Evaluate these Area for 𝛂=25° and length of aquart=1
$$\mathrm{Ratio}\:\mathrm{of}\:\mathrm{green}\:\mathrm{blue}\:\mathrm{and}\:\mathrm{Red}\:\mathrm{triangles}? \\ $$$$\mathrm{Evaluate}\:\mathrm{these}\:\mathrm{Area}\:\mathrm{for}\:\boldsymbol{\alpha}=\mathrm{25}°\:\:\mathrm{and} \\ $$$$\mathrm{length}\:\mathrm{of}\:\mathrm{aquart}=\mathrm{1} \\ $$
Commented by a.lgnaoui last updated on 11/Jan/24
Commented by a.lgnaoui last updated on 11/Jan/24
Answered by Rasheed.Sindhi last updated on 13/Jan/24
Commented by Rasheed.Sindhi last updated on 13/Jan/24
Blue △: cos(25)=(a/c)=(a/1)⇒a=cos(25) sin(25)=(b/c) =(b/1)⇒b=sin(25) Blue▲=((ab)/2)=((cos(25)sin(25))/2) determinant (((Blue▲=((ab)/2)=m((cos(25)sin(25))/2)))) Green △: tan(25)=(d/c)=(d/1)⇒d=tan(25) ▲=(dc/2)=((tan(25)(1))/2)=((tan(25))/2) Green▲=(dc/2)=((tan(25))/2) determinant (((Green▲=(dc/2)=((tan(25))/2)))) Red △: angle between h & g=90−2(25)=40 cos(40)=(h/g)=(h/( (√(c^2 +d^2 ))))=(h/( (√(1+tan^2 (25))))) ⇒h=(√(1+tan^2 (25))) (cos(40)) sin(40)=(i/g)=(i/( (√(c^2 +d^2 )))) =(i/( (√(1+tan^2 (25))))) ⇒i=(√(1+tan^2 (25))) (sin(40)) ▲=((hi)/2)=(((√(1+tan^2 (25))) (cos(40))(√(1+tan^2 (25))) (sin(40)))/2) Red▲=((hi)/2)=(( (cos(40))(sin(40))(1+tan^2 (25)))/2) determinant ((( Red▲=((hi)/2)=(( (cos(40))(sin(40))(1+tan^2 (25)))/2)))) Green▲:Blue▲:Red▲ =((tan(25))/2):((cos(25)sin(25))/2):(( (cos(40))(sin(40))(1+tan^2 (25)))/2) =tan(25):cos(25)sin(25): (cos(40))(sin(40))(1+tan^2 (25))
$$\mathrm{Blue}\:\bigtriangleup: \\ $$$$\:\:\:\mathrm{cos}\left(\mathrm{25}\right)=\frac{{a}}{{c}}=\frac{{a}}{\mathrm{1}}\Rightarrow{a}=\mathrm{cos}\left(\mathrm{25}\right) \\ $$$$\:\:\:\mathrm{sin}\left(\mathrm{25}\right)=\frac{{b}}{{c}}\:=\frac{{b}}{\mathrm{1}}\Rightarrow{b}=\mathrm{sin}\left(\mathrm{25}\right) \\ $$$$\:\:\:\mathrm{Blue}\blacktriangle=\frac{{ab}}{\mathrm{2}}=\frac{\mathrm{cos}\left(\mathrm{25}\right)\mathrm{sin}\left(\mathrm{25}\right)}{\mathrm{2}} \\ $$$$\begin{array}{|c|}{\mathrm{Blue}\blacktriangle=\frac{{ab}}{\mathrm{2}}={m}\frac{\mathrm{cos}\left(\mathrm{25}\right)\mathrm{sin}\left(\mathrm{25}\right)}{\mathrm{2}}}\\\hline\end{array} \\ $$$$\mathrm{Green}\:\bigtriangleup: \\ $$$$\:\:\mathrm{tan}\left(\mathrm{25}\right)=\frac{{d}}{{c}}=\frac{{d}}{\mathrm{1}}\Rightarrow{d}=\mathrm{tan}\left(\mathrm{25}\right) \\ $$$$\:\:\:\blacktriangle=\frac{{dc}}{\mathrm{2}}=\frac{\mathrm{tan}\left(\mathrm{25}\right)\left(\mathrm{1}\right)}{\mathrm{2}}=\frac{\mathrm{tan}\left(\mathrm{25}\right)}{\mathrm{2}} \\ $$$$\:\:\:\mathrm{Green}\blacktriangle=\frac{{dc}}{\mathrm{2}}=\frac{\mathrm{tan}\left(\mathrm{25}\right)}{\mathrm{2}}\: \\ $$$$\:\:\:\begin{array}{|c|}{\mathrm{Green}\blacktriangle=\frac{{dc}}{\mathrm{2}}=\frac{\mathrm{tan}\left(\mathrm{25}\right)}{\mathrm{2}}}\\\hline\end{array} \\ $$$$\mathrm{Red}\:\bigtriangleup: \\ $$$$\:\:\:\:\:{angle}\:{between}\:{h}\:\&\:{g}=\mathrm{90}−\mathrm{2}\left(\mathrm{25}\right)=\mathrm{40} \\ $$$$\:\:\:\:\mathrm{cos}\left(\mathrm{40}\right)=\frac{{h}}{{g}}=\frac{{h}}{\:\sqrt{{c}^{\mathrm{2}} +{d}^{\mathrm{2}} }}=\frac{{h}}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \left(\mathrm{25}\right)}} \\ $$$$\:\:\:\:\:\Rightarrow{h}=\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \left(\mathrm{25}\right)}\:\left(\mathrm{cos}\left(\mathrm{40}\right)\right) \\ $$$$\:\:\:\:\:\mathrm{sin}\left(\mathrm{40}\right)=\frac{{i}}{{g}}=\frac{{i}}{\:\sqrt{{c}^{\mathrm{2}} +{d}^{\mathrm{2}} }}\:=\frac{{i}}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \left(\mathrm{25}\right)}} \\ $$$$\:\:\:\:\:\:\Rightarrow{i}=\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \left(\mathrm{25}\right)}\:\left(\mathrm{sin}\left(\mathrm{40}\right)\right) \\ $$$$\:\:\:\:\:\:\blacktriangle=\frac{{hi}}{\mathrm{2}}=\frac{\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \left(\mathrm{25}\right)}\:\left(\mathrm{cos}\left(\mathrm{40}\right)\right)\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \left(\mathrm{25}\right)}\:\left(\mathrm{sin}\left(\mathrm{40}\right)\right)}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\mathrm{Red}\blacktriangle=\frac{{hi}}{\mathrm{2}}=\frac{\:\left(\mathrm{cos}\left(\mathrm{40}\right)\right)\left(\mathrm{sin}\left(\mathrm{40}\right)\right)\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \left(\mathrm{25}\right)\right)}{\mathrm{2}} \\ $$$$\:\:\:\:\:\begin{array}{|c|}{\:\:\:\:\mathrm{Red}\blacktriangle=\frac{{hi}}{\mathrm{2}}=\frac{\:\left(\mathrm{cos}\left(\mathrm{40}\right)\right)\left(\mathrm{sin}\left(\mathrm{40}\right)\right)\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \left(\mathrm{25}\right)\right)}{\mathrm{2}}}\\\hline\end{array} \\ $$$$\: \\ $$$$\mathrm{Green}\blacktriangle:\mathrm{Blue}\blacktriangle:\mathrm{Red}\blacktriangle \\ $$$$=\frac{\mathrm{tan}\left(\mathrm{25}\right)}{\mathrm{2}}:\frac{\mathrm{cos}\left(\mathrm{25}\right)\mathrm{sin}\left(\mathrm{25}\right)}{\mathrm{2}}:\frac{\:\left(\mathrm{cos}\left(\mathrm{40}\right)\right)\left(\mathrm{sin}\left(\mathrm{40}\right)\right)\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \left(\mathrm{25}\right)\right)}{\mathrm{2}} \\ $$$$=\mathrm{tan}\left(\mathrm{25}\right):\mathrm{cos}\left(\mathrm{25}\right)\mathrm{sin}\left(\mathrm{25}\right):\:\left(\mathrm{cos}\left(\mathrm{40}\right)\right)\left(\mathrm{sin}\left(\mathrm{40}\right)\right)\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \left(\mathrm{25}\right)\right) \\ $$
Commented by a.lgnaoui last updated on 13/Jan/24
Merci
$$\mathrm{Merci}\: \\ $$
Commented by a.lgnaoui last updated on 13/Jan/24
deduire the Area of smal white triangle ?
$$\mathrm{deduire}\:\mathrm{the}\:\mathrm{Area}\:\mathrm{of}\:\mathrm{smal}\:\mathrm{white} \\ $$$$\mathrm{triangle}\:? \\ $$
Commented by Rasheed.Sindhi last updated on 14/Jan/24
Area of whie traingle cos(25)=(e/d)⇒e=dcos(25) sin(25)=(f/d)⇒f=dsin(25) ▲=((ef)/2) =((d^2 cos(25)sin(25) )/2) d=tan(25) =(((tan(25) )^2 cos(25)sin(25) )/2)
$${Area}\:{of}\:{whie}\:{traingle} \\ $$$$\mathrm{cos}\left(\mathrm{25}\right)=\frac{{e}}{{d}}\Rightarrow{e}={d}\mathrm{cos}\left(\mathrm{25}\right)\: \\ $$$$\mathrm{sin}\left(\mathrm{25}\right)=\frac{{f}}{{d}}\Rightarrow{f}={d}\mathrm{sin}\left(\mathrm{25}\right) \\ $$$$\blacktriangle=\frac{{ef}}{\mathrm{2}}\:=\frac{{d}^{\mathrm{2}} \mathrm{cos}\left(\mathrm{25}\right)\mathrm{sin}\left(\mathrm{25}\right)\:\:}{\mathrm{2}}\: \\ $$$${d}=\mathrm{tan}\left(\mathrm{25}\right)\: \\ $$$$\:\:\:\:\:\:\:=\frac{\left(\mathrm{tan}\left(\mathrm{25}\right)\:\right)^{\mathrm{2}} \mathrm{cos}\left(\mathrm{25}\right)\mathrm{sin}\left(\mathrm{25}\right)\:\:}{\mathrm{2}}\: \\ $$$$ \\ $$
Commented by MathematicalUser2357 last updated on 20/Jan/24
tan 25°:(1/2)sin 130°:(1/2)sin 100° sec^2 25° ≈0.46:0.38:0.59
$$\mathrm{tan}\:\mathrm{25}°:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{130}°:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{100}°\:\mathrm{sec}^{\mathrm{2}} \mathrm{25}° \\ $$$$\approx\mathrm{0}.\mathrm{46}:\mathrm{0}.\mathrm{38}:\mathrm{0}.\mathrm{59} \\ $$

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