Question Number 192001 by ajfour last updated on 05/May/23
Commented by ajfour last updated on 05/May/23
$${Find}\:{x},\:{in}\:{terms}\:{of}\:{a},\:{b},\:{m}=\mathrm{tan}\:\theta. \\ $$
Answered by ajfour last updated on 05/May/23
$${say}\:\theta=\mathrm{2}\phi,\:{x}={c} \\ $$$$\mathrm{cos}\:\left(\phi+\mathrm{sin}^{−\mathrm{1}} \frac{{a}−{c}}{{a}+{c}}+\mathrm{sin}^{−\mathrm{1}} \frac{{a}−{b}}{{a}+{b}}+\phi\right) \\ $$$$\:\:\:\:=\frac{\left({a}+{b}\right)^{\mathrm{2}} −\left({b}+{c}\right)^{\mathrm{2}} +\left({c}+{a}\right)^{\mathrm{2}} }{\mathrm{2}\left({a}+{b}\right)\left({c}+{a}\right)} \\ $$$${say}\:{if}\:\phi=\mathrm{0} \\ $$$$\left(\frac{\mathrm{2}\sqrt{{ac}}}{{a}+{c}}\right)\left(\frac{\mathrm{2}\sqrt{{ab}}}{{a}+{b}}\right)−\left(\frac{{a}−{c}}{{a}+{c}}\right)\left(\frac{{a}−{b}}{{a}+{b}}\right) \\ $$$$=\frac{\left({a}+{b}\right)^{\mathrm{2}} −\left({b}+{c}\right)^{\mathrm{2}} +\left({c}+{a}\right)^{\mathrm{2}} }{\mathrm{2}\left({a}+{b}\right)\left({c}+{a}\right)} \\ $$$$\Rightarrow\:\:\mathrm{4}{a}\sqrt{{bc}}−\left\{{a}^{\mathrm{2}} −\left({b}+{c}\right){a}+{bc}\right\} \\ $$$$\:\:\:={a}^{\mathrm{2}} +{a}\left({b}+{c}\right)−{bc} \\ $$$$\Rightarrow\:\:\mathrm{2}{a}^{\mathrm{2}} −\mathrm{4}{a}\sqrt{{bc}}=\mathrm{0} \\ $$$$\Rightarrow\:\:{if}\:{a}\neq\mathrm{0}\:\:\:{then} \\ $$$${a}=\mathrm{2}\sqrt{{bc}} \\ $$$$\checkmark \\ $$
Answered by mr W last updated on 05/May/23
Commented by mr W last updated on 05/May/23
$$\mathrm{cos}\:\alpha=\frac{{a}−{b}}{{a}+{b}} \\ $$$$\mathrm{cos}\:\beta=\frac{{a}−{x}}{{a}+{x}} \\ $$$$\mathrm{cos}\:\gamma=\frac{\left({a}+{b}\right)^{\mathrm{2}} +\left({a}+{x}\right)^{\mathrm{2}} −\left({b}+{x}\right)^{\mathrm{2}} }{\mathrm{2}\left({a}+{b}\right)\left({a}+{x}\right)}=\frac{{a}\left({a}+{b}\right)+\left({a}−{b}\right){x}}{{a}\left({a}+{b}\right)+\left({a}+{b}\right){x}} \\ $$$$\alpha+\beta+\gamma=\mathrm{180}+\theta \\ $$$$\gamma=\mathrm{180}+\theta−\left(\alpha+\beta\right) \\ $$$${let}\:\delta=\alpha−\theta=\mathrm{cos}^{−\mathrm{1}} \frac{{a}−{b}}{{a}+{b}}−\theta \\ $$$$\mathrm{cos}\:\gamma=−\mathrm{cos}\:\left(\delta+\beta\right)=−\mathrm{cos}\:\delta\:\mathrm{cos}\:\beta+\mathrm{sin}\:\delta\:\mathrm{sin}\:\beta \\ $$$$\Rightarrow\frac{{a}\left({a}+{b}\right)+\left({a}−{b}\right){x}}{{a}\left({a}+{b}\right)+\left({a}+{b}\right){x}}+\left(\frac{{a}−{x}}{{a}+{x}}\right)\:\mathrm{cos}\:\delta−\mathrm{sin}\:\delta\:\sqrt{\mathrm{1}−\left(\frac{{a}−{x}}{{a}+{x}}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$