kadetskihykw

2022-05-01

The sides a,b,c of a $\mathrm{\u25b3}ABC$ are in GP whose common ratio is $\frac{2}{3}$ and the circumradius of the triangle is $6\sqrt{\frac{7}{209}}$.Find the longest side of the triangle.

I used law of sines

$\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=2R$

I took $a=a,b=\frac{2}{3}a,c={\left(\frac{2}{3}\right)}^{2}a$

which gives

$\frac{\mathrm{sin}A}{\mathrm{sin}B}=\frac{\mathrm{sin}B}{\mathrm{sin}C}$

I am stuck now,how to find the longest side a

I used law of sines

$\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=2R$

I took $a=a,b=\frac{2}{3}a,c={\left(\frac{2}{3}\right)}^{2}a$

which gives

$\frac{\mathrm{sin}A}{\mathrm{sin}B}=\frac{\mathrm{sin}B}{\mathrm{sin}C}$

I am stuck now,how to find the longest side a

Eliza Flores

Beginner2022-05-02Added 16 answers

If the side lengths are $l,\frac{2}{3}l,\frac{4}{9}l$ then the circumradius is given by:

$R=\frac{abc}{4\mathrm{\Delta}}=\frac{\frac{8}{27}{l}^{3}}{{l}^{2}\sqrt{(1+\frac{2}{3}+\frac{4}{9})(1-\frac{2}{3}+\frac{4}{9})(1+\frac{2}{3}-\frac{4}{9})(-1+\frac{2}{3}+\frac{4}{9})}}$

by Heron's formula, hence:

$R=\frac{24}{\sqrt{1463}}l$

gives:

$l={\frac{7}{4}}.$

$R=\frac{abc}{4\mathrm{\Delta}}=\frac{\frac{8}{27}{l}^{3}}{{l}^{2}\sqrt{(1+\frac{2}{3}+\frac{4}{9})(1-\frac{2}{3}+\frac{4}{9})(1+\frac{2}{3}-\frac{4}{9})(-1+\frac{2}{3}+\frac{4}{9})}}$

by Heron's formula, hence:

$R=\frac{24}{\sqrt{1463}}l$

gives:

$l={\frac{7}{4}}.$

Aliana Kaufman

Beginner2022-05-03Added 13 answers

Let a,b,c be the side lengths. Then, b,c can be written as

$b=\frac{2}{3}a,\phantom{\rule{1em}{0ex}}c={\left(\frac{2}{3}\right)}^{2}a$

where a is the longest side. So, we have, by the law of cosines,

${a}^{2}={\left(\frac{2}{3}a\right)}^{2}+{\left({\left(\frac{2}{3}\right)}^{2}a\right)}^{2}-2\cdot \frac{2}{3}a\cdot {\left(\frac{2}{3}\right)}^{2}a\mathrm{cos}A$

$\Rightarrow \mathrm{cos}A=-\frac{29}{48}.$

Hence,

$a=2R\mathrm{sin}A=2R\sqrt{1-{\mathrm{cos}}^{2}A}=2\cdot 6\sqrt{\frac{7}{209}}\sqrt{1-{(-\frac{29}{48})}^{2}}=\frac{7}{4}.$

$b=\frac{2}{3}a,\phantom{\rule{1em}{0ex}}c={\left(\frac{2}{3}\right)}^{2}a$

where a is the longest side. So, we have, by the law of cosines,

${a}^{2}={\left(\frac{2}{3}a\right)}^{2}+{\left({\left(\frac{2}{3}\right)}^{2}a\right)}^{2}-2\cdot \frac{2}{3}a\cdot {\left(\frac{2}{3}\right)}^{2}a\mathrm{cos}A$

$\Rightarrow \mathrm{cos}A=-\frac{29}{48}.$

Hence,

$a=2R\mathrm{sin}A=2R\sqrt{1-{\mathrm{cos}}^{2}A}=2\cdot 6\sqrt{\frac{7}{209}}\sqrt{1-{(-\frac{29}{48})}^{2}}=\frac{7}{4}.$

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