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Question Number 45980 by ajfour last updated on 19/Oct/18

Commented by ajfour last updated on 19/Oct/18

Determine sides of △,   a, and b = c , in terms of R, r, d.

$${Determine}\:{sides}\:{of}\:\bigtriangleup,\: \\ $$$$\boldsymbol{{a}},\:{and}\:\boldsymbol{{b}}\:=\:\boldsymbol{{c}}\:,\:{in}\:{terms}\:{of}\:\boldsymbol{{R}},\:\boldsymbol{{r}},\:\boldsymbol{{d}}. \\ $$

Answered by MrW3 last updated on 19/Oct/18

let θ=∠B=∠C  cos θ=((R−r)/d)  a=2×(R/(tan (θ/2)))=2R(√((1+cos θ)/(1−cos θ)))  ⇒a=2R(√((1+((R−r)/d))/(1−((R−r)/d))))=2R(√((d+R−r)/(d+r−R)))  ⇒b=c=(a/(2 cos θ))=R(d/(R−r))(√((d+R−r)/(d+r−R)))

$${let}\:\theta=\angle{B}=\angle{C} \\ $$$$\mathrm{cos}\:\theta=\frac{{R}−{r}}{{d}} \\ $$$${a}=\mathrm{2}×\frac{{R}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}=\mathrm{2}{R}\sqrt{\frac{\mathrm{1}+\mathrm{cos}\:\theta}{\mathrm{1}−\mathrm{cos}\:\theta}} \\ $$$$\Rightarrow{a}=\mathrm{2}{R}\sqrt{\frac{\mathrm{1}+\frac{{R}−{r}}{{d}}}{\mathrm{1}−\frac{{R}−{r}}{{d}}}}=\mathrm{2}{R}\sqrt{\frac{{d}+{R}−{r}}{{d}+{r}−{R}}} \\ $$$$\Rightarrow{b}={c}=\frac{{a}}{\mathrm{2}\:\mathrm{cos}\:\theta}={R}\frac{{d}}{{R}−{r}}\sqrt{\frac{{d}+{R}−{r}}{{d}+{r}−{R}}} \\ $$

Commented by ajfour last updated on 19/Oct/18

wonderfully solved, Sir !

$${wonderfully}\:{solved},\:{Sir}\:! \\ $$

Commented by MrW3 last updated on 19/Oct/18

thanks!

$${thanks}! \\ $$

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