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Question-154736

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Question Number 154736 by mathdanisur last updated on 21/Sep/21
Commented by Tawa11 last updated on 21/Sep/21
Weldone sir.
$$\mathrm{Weldone}\:\mathrm{sir}. \\ $$
Answered by mr W last updated on 21/Sep/21
Commented by mr W last updated on 21/Sep/21
((16)/(tan θ))+16+7+(7/(tan ((π/2)−(θ/2))))=((16)/(sin θ))+((16)/(cos θ)) 23+7 tan (θ/2)=16((1/(sin θ))+(1/(cos θ))−(1/(tan θ))) let t=tan (θ/2) 23+7t=16(((1+t^2 )/(2t))+((1+t^2 )/(1−t^2 ))−((1−t^2 )/(2t))) 23−9t=((16(1+t^2 ))/(1−t^2 )) 9t^3 −39t^2 −9t+7=0 (3t−1)(3t^2 −12t−7)=0 t=(1/3) ✓ t=2±((√(51))/3) (rejected, since 0<tan (θ/2)<1) tan (θ/2)=(1/3) tan θ=((2×(1/3))/(1−((1/3))^2 ))=(3/4) A_(blue) =((16)/2)×16 tan θ+(7/2)×(7/(tan (θ/2))) A_(blue) =128×(3/4)+((49)/(2×(1/3))) A_(blue) =((339)/2)=169.5
$$\frac{\mathrm{16}}{\mathrm{tan}\:\theta}+\mathrm{16}+\mathrm{7}+\frac{\mathrm{7}}{\mathrm{tan}\:\left(\frac{\pi}{\mathrm{2}}−\frac{\theta}{\mathrm{2}}\right)}=\frac{\mathrm{16}}{\mathrm{sin}\:\theta}+\frac{\mathrm{16}}{\mathrm{cos}\:\theta} \\ $$$$\mathrm{23}+\mathrm{7}\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}=\mathrm{16}\left(\frac{\mathrm{1}}{\mathrm{sin}\:\theta}+\frac{\mathrm{1}}{\mathrm{cos}\:\theta}−\frac{\mathrm{1}}{\mathrm{tan}\:\theta}\right) \\ $$$${let}\:{t}=\mathrm{tan}\:\frac{\theta}{\mathrm{2}} \\ $$$$\mathrm{23}+\mathrm{7}{t}=\mathrm{16}\left(\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{2}{t}}+\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}} }−\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{2}{t}}\right) \\ $$$$\mathrm{23}−\mathrm{9}{t}=\frac{\mathrm{16}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$$\mathrm{9}{t}^{\mathrm{3}} −\mathrm{39}{t}^{\mathrm{2}} −\mathrm{9}{t}+\mathrm{7}=\mathrm{0} \\ $$$$\left(\mathrm{3}{t}−\mathrm{1}\right)\left(\mathrm{3}{t}^{\mathrm{2}} −\mathrm{12}{t}−\mathrm{7}\right)=\mathrm{0} \\ $$$${t}=\frac{\mathrm{1}}{\mathrm{3}}\:\checkmark \\ $$$${t}=\mathrm{2}\pm\frac{\sqrt{\mathrm{51}}}{\mathrm{3}}\:\left({rejected},\:{since}\:\mathrm{0}<\mathrm{tan}\:\frac{\theta}{\mathrm{2}}<\mathrm{1}\right) \\ $$$$\mathrm{tan}\:\frac{\theta}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{tan}\:\theta=\frac{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} }=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${A}_{{blue}} =\frac{\mathrm{16}}{\mathrm{2}}×\mathrm{16}\:\mathrm{tan}\:\theta+\frac{\mathrm{7}}{\mathrm{2}}×\frac{\mathrm{7}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}} \\ $$$${A}_{{blue}} =\mathrm{128}×\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{49}}{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$${A}_{{blue}} =\frac{\mathrm{339}}{\mathrm{2}}=\mathrm{169}.\mathrm{5} \\ $$
Commented by mathdanisur last updated on 21/Sep/21
Creativ solution Ser, thank you
$$\mathrm{Creativ}\:\mathrm{solution}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{thank}\:\mathrm{you} \\ $$

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