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Question Number 81149 by TawaTawa last updated on 09/Feb/20

Answered by mind is power last updated on 09/Feb/20

tan(B)=((AC)/(AB)) tg(B−θ)=((AD)/(AB))⇒((tg(B))/(tg(B−θ)))=((AC)/(AD))=4 ⇒tg(B)=4tg(B−θ)⇒tg(B)=4((tg(B)−tg(θ))/(1+tg(B)tg(θ))) ⇒3tg(B)=tg(θ)(tg^2 (B)+4) ⇒tg(θ)=((3tg(B))/(4+tg^2 (B)))

$${tan}\left({B}\right)=\frac{{AC}}{{AB}} \\ $$$${tg}\left({B}−\theta\right)=\frac{{AD}}{{AB}}\Rightarrow\frac{{tg}\left({B}\right)}{{tg}\left({B}−\theta\right)}=\frac{{AC}}{{AD}}=\mathrm{4} \\ $$$$\Rightarrow{tg}\left({B}\right)=\mathrm{4}{tg}\left({B}−\theta\right)\Rightarrow{tg}\left({B}\right)=\mathrm{4}\frac{{tg}\left({B}\right)−{tg}\left(\theta\right)}{\mathrm{1}+{tg}\left({B}\right){tg}\left(\theta\right)} \\ $$$$\Rightarrow\mathrm{3}{tg}\left({B}\right)={tg}\left(\theta\right)\left({tg}^{\mathrm{2}} \left({B}\right)+\mathrm{4}\right) \\ $$$$\Rightarrow{tg}\left(\theta\right)=\frac{\mathrm{3}{tg}\left({B}\right)}{\mathrm{4}+{tg}^{\mathrm{2}} \left({B}\right)} \\ $$$$ \\ $$

Commented by TawaTawa last updated on 09/Feb/20

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by mind is power last updated on 09/Feb/20

withe pleasur miss

$${withe}\:{pleasur}\:{miss} \\ $$

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