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Question-187827

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Question Number 187827 by Rupesh123 last updated on 22/Feb/23
Answered by cortano12 last updated on 23/Feb/23
(1)tan 30°=(4/a)⇒a=4(√3) (2) tan 22,5^o =(4/b) ; tan 45°=((2tan 22,5^o )/(1−tan^2 22,5^o )) ⇒tan^2 22,5^o +2tan 22,5^o −1=0 ⇒tan 22,5^o =((−2+2(√2))/2)=(√2)−1 ⇒b=(4/( (√2)−1))=4(√2)+4 (3) tan 37,5^o =(4/c) ; tan 75^o =((2tan 37,5^o )/(1−tan^2 37,5^o )) ⇒2+(√3)=((2tan 37,5^o )/(1−tan^2 37,5^o )) ⇒(2+(√3))tan^2 37,5^o +2tan 37,5^o −(2+(√3))=0 ⇒tan 37,5^o =((−2+(√(11+2(√(12)))))/(2(2+(√3)))) ⇒tan 37,5^o =((5(√3)−8)/2) ⇒c=(8/( 5(√3)−2))=(8/(71))(5(√3)+2) ∴ sides length of the triangle = { ((a+b=4(√3)+4(√2)+4)),((a+c=4(√3)+((16+40(√3))/(71))=((324(√3)+16)/(71)))),((b+c=4(√2)+4+((40(√3)+16)/(71))=((300+284(√2)+40(√3))/(71)))) :}
$$\left(\mathrm{1}\right)\mathrm{tan}\:\mathrm{30}°=\frac{\mathrm{4}}{\mathrm{a}}\Rightarrow\mathrm{a}=\mathrm{4}\sqrt{\mathrm{3}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{tan}\:\mathrm{22},\mathrm{5}^{\mathrm{o}} \:=\frac{\mathrm{4}}{\mathrm{b}}\:;\:\mathrm{tan}\:\mathrm{45}°=\frac{\mathrm{2tan}\:\mathrm{22},\mathrm{5}^{\mathrm{o}} }{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \mathrm{22},\mathrm{5}^{\mathrm{o}} } \\ $$$$\Rightarrow\mathrm{tan}\:^{\mathrm{2}} \mathrm{22},\mathrm{5}^{\mathrm{o}} +\mathrm{2tan}\:\mathrm{22},\mathrm{5}^{\mathrm{o}} −\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{tan}\:\mathrm{22},\mathrm{5}^{\mathrm{o}} =\frac{−\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}}=\sqrt{\mathrm{2}}−\mathrm{1} \\ $$$$\Rightarrow\mathrm{b}=\frac{\mathrm{4}}{\:\sqrt{\mathrm{2}}−\mathrm{1}}=\mathrm{4}\sqrt{\mathrm{2}}+\mathrm{4} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{tan}\:\mathrm{37},\mathrm{5}^{\mathrm{o}} =\frac{\mathrm{4}}{\mathrm{c}}\:;\:\mathrm{tan}\:\mathrm{75}^{\mathrm{o}} =\frac{\mathrm{2tan}\:\mathrm{37},\mathrm{5}^{\mathrm{o}} }{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\mathrm{37},\mathrm{5}^{\mathrm{o}} } \\ $$$$\Rightarrow\mathrm{2}+\sqrt{\mathrm{3}}=\frac{\mathrm{2tan}\:\mathrm{37},\mathrm{5}^{\mathrm{o}} }{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \mathrm{37},\mathrm{5}^{\mathrm{o}} } \\ $$$$\Rightarrow\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\mathrm{tan}\:^{\mathrm{2}} \mathrm{37},\mathrm{5}^{\mathrm{o}} +\mathrm{2tan}\:\mathrm{37},\mathrm{5}^{\mathrm{o}} −\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{tan}\:\mathrm{37},\mathrm{5}^{\mathrm{o}} =\frac{−\mathrm{2}+\sqrt{\mathrm{11}+\mathrm{2}\sqrt{\mathrm{12}}}}{\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)} \\ $$$$\Rightarrow\mathrm{tan}\:\mathrm{37},\mathrm{5}^{\mathrm{o}} =\frac{\mathrm{5}\sqrt{\mathrm{3}}−\mathrm{8}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{c}=\frac{\mathrm{8}}{\:\mathrm{5}\sqrt{\mathrm{3}}−\mathrm{2}}=\frac{\mathrm{8}}{\mathrm{71}}\left(\mathrm{5}\sqrt{\mathrm{3}}+\mathrm{2}\right) \\ $$$$\therefore\:\mathrm{sides}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle} \\ $$$$\:=\begin{cases}{\mathrm{a}+\mathrm{b}=\mathrm{4}\sqrt{\mathrm{3}}+\mathrm{4}\sqrt{\mathrm{2}}+\mathrm{4}}\\{\mathrm{a}+\mathrm{c}=\mathrm{4}\sqrt{\mathrm{3}}+\frac{\mathrm{16}+\mathrm{40}\sqrt{\mathrm{3}}}{\mathrm{71}}=\frac{\mathrm{324}\sqrt{\mathrm{3}}+\mathrm{16}}{\mathrm{71}}}\\{\mathrm{b}+\mathrm{c}=\mathrm{4}\sqrt{\mathrm{2}}+\mathrm{4}+\frac{\mathrm{40}\sqrt{\mathrm{3}}+\mathrm{16}}{\mathrm{71}}=\frac{\mathrm{300}+\mathrm{284}\sqrt{\mathrm{2}}+\mathrm{40}\sqrt{\mathrm{3}}}{\mathrm{71}}}\end{cases}\:\:\: \\ $$$$ \\ $$
Commented by Rupesh123 last updated on 22/Feb/23
Very nice work!
Answered by nikif99 last updated on 22/Feb/23
Let incenter: I. Let ID radius=4 cm ∡C=75° AD=((ID)/(tan 22.5))=4((√2)+1) BD=((ID)/(tan 30))=4(√3) AB=4(1+(√2)+(√3)) cm (1) ((AC)/(sin 60))=((AB)/(sin 75))=((BC)/(sin 45)) (2) (1)(2) ⇒AC=4(3−(√3)+(√6)) cm BC=4(2−(√2)+(√6)) cm
$${Let}\:{incenter}:\:{I}.\:{Let}\:{ID}\:{radius}=\mathrm{4}\:{cm} \\ $$$$\measuredangle{C}=\mathrm{75}° \\ $$$${AD}=\frac{{ID}}{\mathrm{tan}\:\mathrm{22}.\mathrm{5}}=\mathrm{4}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right) \\ $$$${BD}=\frac{{ID}}{\mathrm{tan}\:\mathrm{30}}=\mathrm{4}\sqrt{\mathrm{3}} \\ $$$${AB}=\mathrm{4}\left(\mathrm{1}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}\right)\:{cm}\:\left(\mathrm{1}\right) \\ $$$$\frac{{AC}}{\mathrm{sin}\:\mathrm{60}}=\frac{{AB}}{\mathrm{sin}\:\mathrm{75}}=\frac{{BC}}{\mathrm{sin}\:\mathrm{45}}\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)\left(\mathrm{2}\right)\:\Rightarrow{AC}=\mathrm{4}\left(\mathrm{3}−\sqrt{\mathrm{3}}+\sqrt{\mathrm{6}}\right)\:{cm} \\ $$$${BC}=\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{2}}+\sqrt{\mathrm{6}}\right)\:{cm} \\ $$

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