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Question-70887




Question Number 70887 by naka3546 last updated on 09/Oct/19
Commented by ajfour last updated on 09/Oct/19
Commented by ajfour last updated on 09/Oct/19
ssin α=rsin γ  b^2 =(ssin α+rcos δ)^2 +(scos α−rsin δ)^2   rsin δ=asin β  b^2 =(rsin δ+asin β)^2 +(acos β−rcos δ)^2   δ+γ=(π/2)  ((scos α−rsin δ)/(ssin α+rcos δ))=((rsin δ+asin β)/(acos β−rcos δ))  .........
$${s}\mathrm{sin}\:\alpha={r}\mathrm{sin}\:\gamma \\ $$$${b}^{\mathrm{2}} =\left({s}\mathrm{sin}\:\alpha+{r}\mathrm{cos}\:\delta\right)^{\mathrm{2}} +\left({s}\mathrm{cos}\:\alpha−{r}\mathrm{sin}\:\delta\right)^{\mathrm{2}} \\ $$$${r}\mathrm{sin}\:\delta={a}\mathrm{sin}\:\beta \\ $$$${b}^{\mathrm{2}} =\left({r}\mathrm{sin}\:\delta+{a}\mathrm{sin}\:\beta\right)^{\mathrm{2}} +\left({a}\mathrm{cos}\:\beta−{r}\mathrm{cos}\:\delta\right)^{\mathrm{2}} \\ $$$$\delta+\gamma=\frac{\pi}{\mathrm{2}} \\ $$$$\frac{{s}\mathrm{cos}\:\alpha−{r}\mathrm{sin}\:\delta}{{s}\mathrm{sin}\:\alpha+{r}\mathrm{cos}\:\delta}=\frac{{r}\mathrm{sin}\:\delta+{a}\mathrm{sin}\:\beta}{{a}\mathrm{cos}\:\beta−{r}\mathrm{cos}\:\delta} \\ $$$$……… \\ $$

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