Question and Answers Forum

All Questions      Topic List

Mensuration Questions

Previous in All Question      Next in All Question      

Previous in Mensuration      Next in Mensuration      

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

$${Let}\:{incenter}:\:{I}.\:{Let}\:{ID}\:{radius}=\mathrm{4}\:{cm} \\ $$$$\measuredangle{C}=\mathrm{75}° \\ $$$${AD}=\frac{{ID}}{\mathrm{tan}\:\mathrm{22}.\mathrm{5}}=\mathrm{4}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right) \\ $$$${BD}=\frac{{ID}}{\mathrm{tan}\:\mathrm{30}}=\mathrm{4}\sqrt{\mathrm{3}} \\ $$$${AB}=\mathrm{4}\left(\mathrm{1}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}\right)\:{cm}\:\left(\mathrm{1}\right) \\ $$$$\frac{{AC}}{\mathrm{sin}\:\mathrm{60}}=\frac{{AB}}{\mathrm{sin}\:\mathrm{75}}=\frac{{BC}}{\mathrm{sin}\:\mathrm{45}}\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)\left(\mathrm{2}\right)\:\Rightarrow{AC}=\mathrm{4}\left(\mathrm{3}−\sqrt{\mathrm{3}}+\sqrt{\mathrm{6}}\right)\:{cm} \\ $$$${BC}=\mathrm{4}\left(\mathrm{2}−\sqrt{\mathrm{2}}+\sqrt{\mathrm{6}}\right)\:{cm} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com