Question Number 19675 by ajfour last updated on 14/Aug/17
Commented by ajfour last updated on 14/Aug/17
$$\mathrm{Q}.\mathrm{19668}\:\mathrm{Find}\:\mathrm{r}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{side}\:\boldsymbol{\mathrm{d}} \\ $$$$\mathrm{of}\:\mathrm{equilareral}\:\bigtriangleup\mathrm{ABC}\:. \\ $$
Answered by ajfour last updated on 14/Aug/17
$$\:\:\frac{\mathrm{r}}{\mathrm{x}}=\mathrm{sin}\:\left(\mathrm{90}−\theta\right)=\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{x}=\mathrm{2r} \\ $$$$\mathrm{x}+\mathrm{r}+\mathrm{R}=\frac{\mathrm{d}}{\mathrm{sin}\:\theta}\:\:\left(\mathrm{C}\:\mathrm{to}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{big}\:\mathrm{circle}\right) \\ $$$$\Rightarrow\:\:\mathrm{3r}+\frac{\mathrm{d}}{\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{2d}}{\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow\:\:\mathrm{r}=\frac{\mathrm{d}}{\mathrm{3}\sqrt{\mathrm{3}}}\:=\:\frac{\boldsymbol{\mathrm{d}}\sqrt{\mathrm{3}}}{\mathrm{9}}\:.\:\:\left(\mathrm{E}\right) \\ $$
Commented by Joel577 last updated on 15/Aug/17
$${thank}\:{you}\:{very}\:{much} \\ $$