Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 33466 by mondodotto@gmail.com last updated on 17/Apr/18

$$\mathbf{if}\mathbf{tan}\mathit{\beta }=\frac{\mathbf{r}}{\sqrt{\mathbf{s}}}\mathbf{and}\mathbf{sin}\mathit{\beta }=\frac{\sqrt{\mathbf{s}}}{\mathbf{r}}$$$$\mathbf{show}\mathbf{that}\mathbf{cos}\mathit{\beta }=\sqrt{{\mathbf{r}}^2+\mathbf{s}}$$

Commented by MJS last updated on 17/Apr/18

$$\cos \beta =\sqrt{1-{\sin }^2\beta }=\frac{\sqrt{r^2-s}}{\mid r\mid }$$$$\sqrt{r^2+s}=\frac{\sqrt{r^2-s}}{\mid r\mid }\Rightarrow s=r^2\frac{1-r^2}{1+r^2}$$$$$$$$\cos \beta =\frac{\sin \beta }{\tan \beta }$$$$\sqrt{r^2+s}=\frac{s}{r^2}\Rightarrow s=\frac{r^3}{2}(r\pm \sqrt{4+r^2})$$$$$$$$r^2\frac{1-r^2}{1+r^2}=\frac{r^3}{2}(r\pm \sqrt{4+r^2})\Rightarrow$$$$\Rightarrow r=\sqrt{-2+\sqrt{5}};s=\frac{7}{2}-\frac{3}{2}\sqrt{5}$$$$\beta \approx 51.83°$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{only}\mathrm{true}\mathrm{in}\mathrm{this}\mathrm{singular}\mathrm{case}$$

Commented by mondodotto@gmail.com last updated on 17/Apr/18

$$\mathrm{thanx}\mathrm{for}\mathrm{your}\mathrm{help}\mathrm{sir}$$

Answered by Rasheed.Sindhi last updated on 17/Apr/18

$$\mathbf{tan}\mathit{\beta }=\frac{\mathbf{sin}\mathit{\beta }}{\mathbf{cos}\mathit{\beta }}$$$$$$$$\frac{\mathbf{r}}{\sqrt{\mathbf{s}}}=\frac{\frac{\sqrt{\mathbf{s}}}{\mathbf{r}}}{\mathbf{cos}\mathit{\beta }}$$$$$$$$\mathbf{cos}\mathit{\beta }=\frac{\frac{\sqrt{\mathbf{s}}}{\mathbf{r}}}{\frac{\mathbf{r}}{\sqrt{\mathbf{s}}}}=\frac{\sqrt{\mathbf{s}}}{\mathbf{r}}\times \frac{\sqrt{\mathrm{s}}}{\mathrm{r}}=\frac{\mathrm{s}}{{\mathrm{r}}^2}\ne \sqrt{{\mathbf{r}}^2+\mathbf{s}}\mathrm{in}\mathrm{general}.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com