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 179938 by Acem last updated on 04/Nov/22

Commented by Acem last updated on 04/Nov/22

It′s correct, Thanks Sir!

$${It}'{s}\:{correct},\:{Thanks}\:{Sir}! \\ $$

Commented by HeferH last updated on 04/Nov/22

Commented by HeferH last updated on 04/Nov/22

x(√3) − x = 1 x = (1/( (√3)−1)) = (((√3) + 1)/2) tan β = ((x(√3))/(2 − x)) = ((3 + (√3))/(3− (√3))) = 2 + (√3) β = tan^(−1) (2 + (√3)) β = 75 °

$${x}\sqrt{\mathrm{3}}\:−\:{x}\:=\:\mathrm{1} \\ $$$$\:{x}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}−\mathrm{1}}\:=\:\frac{\sqrt{\mathrm{3}}\:+\:\mathrm{1}}{\mathrm{2}} \\ $$$$\:\mathrm{tan}\:\beta\:\:=\:\frac{{x}\sqrt{\mathrm{3}}}{\mathrm{2}\:−\:{x}}\:=\:\frac{\mathrm{3}\:+\:\sqrt{\mathrm{3}}}{\mathrm{3}−\:\sqrt{\mathrm{3}}}\:=\:\mathrm{2}\:+\:\sqrt{\mathrm{3}} \\ $$$$\:\beta\:=\:\mathrm{tan}^{−\mathrm{1}} \:\left(\mathrm{2}\:+\:\sqrt{\mathrm{3}}\right)\: \\ $$$$\:\beta\:=\:\mathrm{75}\:° \\ $$

Answered by mr W last updated on 04/Nov/22

((2 sin β)/(sin (β+60)))=((1×sin 45)/(sin 15)) ((sin β)/(sin (β+60)))=(4/( 2(√2) ((√6)−(√2)))) ((sin 60)/(tan β))+cos 60=(√3)−1 tan β=2+(√3) ⇒β=75°

$$\frac{\mathrm{2}\:\mathrm{sin}\:\beta}{\mathrm{sin}\:\left(\beta+\mathrm{60}\right)}=\frac{\mathrm{1}×\mathrm{sin}\:\mathrm{45}}{\mathrm{sin}\:\mathrm{15}} \\ $$$$\frac{\mathrm{sin}\:\beta}{\mathrm{sin}\:\left(\beta+\mathrm{60}\right)}=\frac{\mathrm{4}}{\:\mathrm{2}\sqrt{\mathrm{2}}\:\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}\right)} \\ $$$$\frac{\mathrm{sin}\:\mathrm{60}}{\mathrm{tan}\:\beta}+\mathrm{cos}\:\mathrm{60}=\sqrt{\mathrm{3}}−\mathrm{1} \\ $$$$\mathrm{tan}\:\beta=\mathrm{2}+\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\beta=\mathrm{75}° \\ $$

Commented by Acem last updated on 05/Nov/22

Very good Sir! thanks

$${Very}\:{good}\:{Sir}!\:{thanks} \\ $$

Answered by Acem last updated on 05/Nov/22

Commented by Acem last updated on 05/Nov/22

λ= 60−45= 15 MN= 2 cos 60= 1 ⇒ MN=MA ⇒ α= φ= 30 ⇒Υ= λ ⇒ NA= NC α= β_1 ⇒ NA= NB= NC ⇒ θ= β_2 = 45 ⇒ 𝛃= 75°

$$\lambda=\:\mathrm{60}−\mathrm{45}=\:\mathrm{15} \\ $$$${MN}=\:\mathrm{2}\:\mathrm{cos}\:\mathrm{60}=\:\mathrm{1}\:\Rightarrow\:{MN}={MA}\:\Rightarrow\:\alpha=\:\phi=\:\mathrm{30} \\ $$$$\:\Rightarrow\Upsilon=\:\lambda\:\Rightarrow\:{NA}=\:{NC} \\ $$$$\:\alpha=\:\beta_{\mathrm{1}} \:\:\:\:\:\Rightarrow\:{NA}=\:{NB}=\:{NC}\:\Rightarrow\:\theta=\:\beta_{\mathrm{2}} =\:\mathrm{45} \\ $$$$\:\Rightarrow\:\boldsymbol{\beta}=\:\mathrm{75}° \\ $$

Commented by mr W last updated on 05/Nov/22

nice solution!

$${nice}\:{solution}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com