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Question Number 172628 by mnjuly1970 last updated on 29/Jun/22

Commented by mr W last updated on 29/Jun/22

maybe (2/(tan γ))=(1/(tan β))−(1/(tan α)) ?

$${maybe}\:\frac{\mathrm{2}}{\mathrm{tan}\:\gamma}=\frac{\mathrm{1}}{\mathrm{tan}\:\beta}−\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}\:? \\ $$

Commented by mnjuly1970 last updated on 29/Jun/22

yes sir W ...thanks alot

$$\:\:\mathrm{yes}\:\:\mathrm{sir}\:\:\mathrm{W}\:...\mathrm{thanks}\:\mathrm{alot} \\ $$

Answered by mr W last updated on 29/Jun/22

((BM)/(sin α))=((AM)/(sin (γ+α))) ...(i) ((sin β)/(MC))=((sin (γ−β))/(AM)) ...(ii) (i)×(ii): ((sin β)/(sin α))=((sin (γ−β))/(sin (γ+α)))=((sin γ cos β−cos γ sin β)/(sin γ cos α+cos γ sin α)) 1=((((tan γ)/(tan β))−1)/(((tan γ)/(tan α))+1)) ⇒(2/(tan γ))=(1/(tan β))−(1/(tan α))

$$\frac{{BM}}{\mathrm{sin}\:\alpha}=\frac{{AM}}{\mathrm{sin}\:\left(\gamma+\alpha\right)}\:\:\:...\left({i}\right) \\ $$$$\frac{\mathrm{sin}\:\beta}{{MC}}=\frac{\mathrm{sin}\:\left(\gamma−\beta\right)}{{AM}}\:\:\:...\left({ii}\right) \\ $$$$\left({i}\right)×\left({ii}\right): \\ $$$$\frac{\mathrm{sin}\:\beta}{\mathrm{sin}\:\alpha}=\frac{\mathrm{sin}\:\left(\gamma−\beta\right)}{\mathrm{sin}\:\left(\gamma+\alpha\right)}=\frac{\mathrm{sin}\:\gamma\:\mathrm{cos}\:\beta−\mathrm{cos}\:\gamma\:\mathrm{sin}\:\beta}{\mathrm{sin}\:\gamma\:\mathrm{cos}\:\alpha+\mathrm{cos}\:\gamma\:\mathrm{sin}\:\alpha} \\ $$$$\mathrm{1}=\frac{\frac{\mathrm{tan}\:\gamma}{\mathrm{tan}\:\beta}−\mathrm{1}}{\frac{\mathrm{tan}\:\gamma}{\mathrm{tan}\:\alpha}+\mathrm{1}} \\ $$$$\Rightarrow\frac{\mathrm{2}}{\mathrm{tan}\:\gamma}=\frac{\mathrm{1}}{\mathrm{tan}\:\beta}−\frac{\mathrm{1}}{\mathrm{tan}\:\alpha} \\ $$

Commented by Tawa11 last updated on 30/Jun/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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