Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 12928 by tawa last updated on 07/May/17

Answered by sandy_suhendra last updated on 09/May/17

cosine rule in ΔABC :  x^2 =d^2 +d^2 −2d.d.cos(180−2θ)  x^2 =2d^2 −2d^2 (−cos2θ) ⇒ use cos2θ=2cos^2 θ−1       x^2 =2d^2 −2d^2 (−2cos^2 θ+1)  x^2 =2d^2 +4d^2 cos^2 θ−2d^2   x^2 =4d^2 cos^2 θ  x=2d cos θ    sine rule in ΔBCD :  (k/(sin θ)) = (x/(sin[180−(θ+β)]))  (k/(sin θ)) = ((2d cos θ)/(sin(θ+β)))  k = ((2d cos θ sin θ)/(sin(θ+β))) ⇒ use sin 2θ=2sinθcosθ       k = ((2d(0.5 sin 2θ))/(sin(α+β)))  k = ((d sin 2θ)/(sin(α+β)))

$$\mathrm{cosine}\:\mathrm{rule}\:\mathrm{in}\:\Delta\mathrm{ABC}\:: \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{d}^{\mathrm{2}} +\mathrm{d}^{\mathrm{2}} −\mathrm{2d}.\mathrm{d}.\mathrm{cos}\left(\mathrm{180}−\mathrm{2}\theta\right) \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{2d}^{\mathrm{2}} −\mathrm{2d}^{\mathrm{2}} \left(−\mathrm{cos2}\theta\right)\:\Rightarrow\:\mathrm{use}\:\mathrm{cos2}\theta=\mathrm{2cos}^{\mathrm{2}} \theta−\mathrm{1}\:\:\:\:\: \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{2d}^{\mathrm{2}} −\mathrm{2d}^{\mathrm{2}} \left(−\mathrm{2cos}^{\mathrm{2}} \theta+\mathrm{1}\right) \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{2d}^{\mathrm{2}} +\mathrm{4d}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \theta−\mathrm{2d}^{\mathrm{2}} \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{4d}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \theta \\ $$$$\mathrm{x}=\mathrm{2d}\:\mathrm{cos}\:\theta \\ $$$$ \\ $$$$\mathrm{sine}\:\mathrm{rule}\:\mathrm{in}\:\Delta\mathrm{BCD}\:: \\ $$$$\frac{\mathrm{k}}{\mathrm{sin}\:\theta}\:=\:\frac{\mathrm{x}}{\mathrm{sin}\left[\mathrm{180}−\left(\theta+\beta\right)\right]} \\ $$$$\frac{\mathrm{k}}{\mathrm{sin}\:\theta}\:=\:\frac{\mathrm{2d}\:\mathrm{cos}\:\theta}{\mathrm{sin}\left(\theta+\beta\right)} \\ $$$$\mathrm{k}\:=\:\frac{\mathrm{2d}\:\mathrm{cos}\:\theta\:\mathrm{sin}\:\theta}{\mathrm{sin}\left(\theta+\beta\right)}\:\Rightarrow\:\mathrm{use}\:\mathrm{sin}\:\mathrm{2}\theta=\mathrm{2sin}\theta\mathrm{cos}\theta\:\:\:\:\: \\ $$$$\mathrm{k}\:=\:\frac{\mathrm{2d}\left(\mathrm{0}.\mathrm{5}\:\mathrm{sin}\:\mathrm{2}\theta\right)}{\mathrm{sin}\left(\alpha+\beta\right)} \\ $$$$\mathrm{k}\:=\:\frac{\mathrm{d}\:\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{sin}\left(\alpha+\beta\right)} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com