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Question Number 11868 by Nayon last updated on 03/Apr/17

Answered by sandy_suhendra last updated on 03/Apr/17

Commented by sandy_suhendra last updated on 03/Apr/17

$$\mathrm{\Delta }\mathrm{FBA}\approx \mathrm{\Delta }{\mathrm{FB}}^{\prime }{\mathrm{A}}^{\prime }(\mathrm{their}\mathrm{corresponding}\mathrm{angle}$$$$\mathrm{have}\mathrm{the}\mathrm{same}\mathrm{measure})$$$$\mathrm{so}\frac{{\mathrm{A}}^{\prime }{\mathrm{O}}^{\prime }}{\mathrm{AO}}=\frac{{\mathrm{A}}^{\prime }{\mathrm{B}}^{\prime }}{\mathrm{AB}}\mathrm{or}\frac{{\mathrm{A}}^{\prime }{\mathrm{O}}^{\prime }}{\mathrm{AO}}=\frac{{\mathrm{O}}^{\prime }\mathrm{P}}{\mathrm{OP}}$$$$$$$$\mathrm{\Delta }{\mathrm{FO}}^{\prime }{\mathrm{A}}^{\prime }\approx \mathrm{\Delta }\mathrm{BFP}$$$$\mathrm{so}\frac{{\mathrm{O}}^{\prime }{\mathrm{A}}^{\prime }}{\mathrm{BP}}=\frac{{\mathrm{O}}^{\prime }\mathrm{F}}{\mathrm{PF}}(\mathrm{BP}\approx \mathrm{OA})$$$$\frac{{\mathrm{O}}^{\prime }{\mathrm{A}}^{\prime }}{\mathrm{OA}}=\frac{{\mathrm{O}}^{\prime }\mathrm{P}-\mathrm{PF}}{\mathrm{PF}}$$$$\frac{{\mathrm{O}}^{\prime }\mathrm{P}}{\mathrm{OP}}=\frac{{\mathrm{O}}^{\prime }\mathrm{P}-\mathrm{PF}}{\mathrm{PF}}$$$$\mathrm{let}{\mathrm{O}}^{\prime }\mathrm{P}={\mathrm{s}}^{\prime }$$$$\mathrm{OP}=\mathrm{s}$$$$\mathrm{PF}=\mathrm{f}$$$$\frac{{\mathrm{s}}^{\prime }}{\mathrm{s}}=\frac{{\mathrm{s}}^{\prime }-\mathrm{f}}{\mathrm{f}}$$$$\mathrm{s}.{\mathrm{s}}^{\prime }-\mathrm{s}.\mathrm{f}={\mathrm{s}}^{\prime }\mathrm{f}\Rightarrow \mathrm{divided}\mathrm{by}\mathrm{s}.{\mathrm{s}}^{\prime }.\mathrm{f}$$$$\frac{\mathrm{s}.{\mathrm{s}}^{\prime }}{\mathrm{s}.{\mathrm{s}}^{\prime }.\mathrm{f}}-\frac{\mathrm{s}.\mathrm{f}}{\mathrm{s}.{\mathrm{s}}^{\prime }.\mathrm{f}}=\frac{{\mathrm{s}}^{\prime }.\mathrm{f}}{\mathrm{s}.{\mathrm{s}}^{\prime }.\mathrm{f}}$$$$\frac{1}{\mathrm{f}}-\frac{1}{{\mathrm{s}}^{\prime }}=\frac{1}{\mathrm{s}}$$$$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{s}}+\frac{1}{{\mathrm{s}}^{\prime }}$$$$\mathrm{or}\frac{1}{\mathrm{PF}}=\frac{1}{\mathrm{OP}}+\frac{1}{{\mathrm{O}}^{\prime }\mathrm{P}}$$

Commented by Nayon last updated on 06/Apr/17

$$hydoodAB\ne OP$$

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