Question-201989

Question Number 201989 by cortano12 last updated on 18/Dec/23

Answered by mr W last updated on 18/Dec/23

Commented by mr W last updated on 18/Dec/23

sin α=(5/(15))=(1/3) sin β=((10)/(20))=(1/2) ⇒β=30° BS^2 =10×30 ⇒BS=10(√3) PB^2 =30^2 +(10(√3))^2 −2×30×10(√3)×((√3)/2)=300 ⇒PB=10(√3)=BS ⇒α+γ=β=30° ((AB)/(sin γ))=((PB)/(sin (α+β))) ⇒AB=((10(√3) sin (30°−α))/(sin (30°+α))) =((10(√3)(cos α−(√3) sin α))/(cos α+(√3) sin α)) =((10(√3)(((2(√2))/3)−((√3)/3)))/(((2(√2))/3)+((√3)/3)))=2(11(√3)−12(√2))≈4.164

$$\mathrm{sin}\:\alpha=\frac{\mathrm{5}}{\mathrm{15}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{sin}\:\beta=\frac{\mathrm{10}}{\mathrm{20}}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\beta=\mathrm{30}° \\ $$$${BS}^{\mathrm{2}} =\mathrm{10}×\mathrm{30}\:\Rightarrow{BS}=\mathrm{10}\sqrt{\mathrm{3}} \\ $$$${PB}^{\mathrm{2}} =\mathrm{30}^{\mathrm{2}} +\left(\mathrm{10}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} −\mathrm{2}×\mathrm{30}×\mathrm{10}\sqrt{\mathrm{3}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{300} \\ $$$$\Rightarrow{PB}=\mathrm{10}\sqrt{\mathrm{3}}={BS}\:\Rightarrow\alpha+\gamma=\beta=\mathrm{30}° \\ $$$$\frac{{AB}}{\mathrm{sin}\:\gamma}=\frac{{PB}}{\mathrm{sin}\:\left(\alpha+\beta\right)} \\ $$$$\Rightarrow{AB}=\frac{\mathrm{10}\sqrt{\mathrm{3}}\:\mathrm{sin}\:\left(\mathrm{30}°−\alpha\right)}{\mathrm{sin}\:\left(\mathrm{30}°+\alpha\right)} \\ $$$$\:\:\:=\frac{\mathrm{10}\sqrt{\mathrm{3}}\left(\mathrm{cos}\:\alpha−\sqrt{\mathrm{3}}\:\mathrm{sin}\:\alpha\right)}{\mathrm{cos}\:\alpha+\sqrt{\mathrm{3}}\:\mathrm{sin}\:\alpha} \\ $$$$\:\:\:=\frac{\mathrm{10}\sqrt{\mathrm{3}}\left(\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\right)}{\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}}=\mathrm{2}\left(\mathrm{11}\sqrt{\mathrm{3}}−\mathrm{12}\sqrt{\mathrm{2}}\right)\approx\mathrm{4}.\mathrm{164} \\ $$

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