Find-tan-

Question Number 212214 by CrispyXYZ last updated on 06/Oct/24

$$\mathrm{Find}\:\mathrm{tan}\theta. \\ $$

Commented by CrispyXYZ last updated on 06/Oct/24

Answered by a.lgnaoui last updated on 06/Oct/24

sin 𝛉=((CD)/(BC)) BC=2(√2) CD=2(√2) sin 𝛉 AD^2 =1+CD^2 −2CDcos (180−𝛉) 9=1+8sin^2 𝛉+2(√2) sin 2𝛉 8=8sin^2 𝛉+2(√2) sin 2𝛉 sin 2𝛉=2(√2) (1−sin^2 𝛉) sin 𝛉(√(1−sin^2 𝛉))=(√2) (1−sin^2 𝛉) sin 𝛉=(√(2(1−sin^2 𝛉))) 3sin^2 𝛉−2=0 ⇒ sin 𝛉=(√(2/3)) ;(𝛉<90) 𝛉=54,735

$$\mathrm{sin}\:\boldsymbol{\theta}=\frac{\mathrm{CD}}{\mathrm{BC}}\:\:\:\:\:\:\mathrm{BC}=\mathrm{2}\sqrt{\mathrm{2}}\:\:\:\:\mathrm{CD}=\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{sin}\:\boldsymbol{\theta} \\ $$$$\:\mathrm{AD}^{\mathrm{2}} =\mathrm{1}+\mathrm{CD}^{\mathrm{2}} −\mathrm{2CDcos}\:\left(\mathrm{180}−\boldsymbol{\theta}\right) \\ $$$$\:\mathrm{9}=\mathrm{1}+\mathrm{8sin}\:^{\mathrm{2}} \boldsymbol{\theta}+\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{sin}\:\mathrm{2}\boldsymbol{\theta} \\ $$$$\mathrm{8}=\mathrm{8sin}\:^{\mathrm{2}} \boldsymbol{\theta}+\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{sin}\:\mathrm{2}\boldsymbol{\theta} \\ $$$$\:\:\:\mathrm{sin}\:\mathrm{2}\boldsymbol{\theta}=\mathrm{2}\sqrt{\mathrm{2}}\:\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \boldsymbol{\theta}\right) \\ $$$$\mathrm{sin}\:\boldsymbol{\theta}\sqrt{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \boldsymbol{\theta}}=\sqrt{\mathrm{2}}\:\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \boldsymbol{\theta}\right) \\ $$$$\mathrm{sin}\:\boldsymbol{\theta}=\sqrt{\mathrm{2}\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \boldsymbol{\theta}\right)} \\ $$$$\:\:\:\mathrm{3sin}\:^{\mathrm{2}} \boldsymbol{\theta}−\mathrm{2}=\mathrm{0}\:\Rightarrow\:\:\mathrm{sin}\:\boldsymbol{\theta}=\sqrt{\frac{\mathrm{2}}{\mathrm{3}}}\:\:;\left(\boldsymbol{\theta}<\mathrm{90}\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\theta}=\mathrm{54},\mathrm{735} \\ $$$$\:\: \\ $$

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