Question Number 60043 by Kunal12588 last updated on 17/May/19
$${Show}\:{that}\:{in}\:{a}\:\mathrm{30}°−\mathrm{60}°−\mathrm{90}°\:{triangle}\:{the}\: \\ $$$${altitude}\:{on}\:{the}\:{hypotaneuse}\:{divides}\:{the}\: \\ $$$${hypotaneuse}\:{into}\:{segments}\:{whose}\:{length} \\ $$$${has}\:{the}\:{ratio}\:\mathrm{1}/\mathrm{3}. \\ $$$${without}\:{using}\:{trigonometry}. \\ $$
Answered by tanmay last updated on 17/May/19
$${The}\:{eqn}\:{of}\:{hypotaneous} \\ $$$$\frac{{x}}{{a}}+\frac{{y}}{{b}}=\mathrm{1} \\ $$$${bx}+{ay}={ab} \\ $$$${y}=\frac{−{bx}}{{a}}+{b}\:\rightarrow{slopse}\:{m}_{\mathrm{1}} =−\frac{{b}}{{a}} \\ $$$${eq}\:{of}\:{st}\:{line}\:\bot{to}\:{hypotaneous}\:{and}\:{passing} \\ $$$${through}\:{origin}\left(\mathrm{0},\mathrm{0}\right) \\ $$$${y}={m}_{\mathrm{2}} {x}\:\:\:\:{m}_{\mathrm{1}} ×{m}_{\mathrm{2}} =−\mathrm{1} \\ $$$${m}_{\mathrm{2}} =\frac{−\mathrm{1}}{\frac{−{b}}{{a}}}=\frac{{a}}{{b}}\:\:\: \\ $$$${y}=\frac{{a}}{{b}}{x} \\ $$$${solve}\:{y}=\frac{{a}}{{b}}{x}\:{and}\:\frac{{x}}{{a}}+\frac{{y}}{{b}}=\mathrm{1} \\ $$$${to}\:{get}\:{point}\:{C}\:{which}\:{devide}\:{AB}\:\:\:{in}\:\frac{\mathrm{1}}{\mathrm{3}}\:\:{ratio} \\ $$$$\frac{{x}}{{b}}=\frac{{y}}{{a}}={k}\rightarrow{x}={bk}\:\:\:{y}={ak}\:\: \\ $$$$\frac{{bk}}{{a}}+\frac{{ak}}{{b}}=\mathrm{1}\rightarrow{k}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)={ab}\:\:{k}=\frac{{ab}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${x}=\frac{{ab}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:\:\:{y}=\frac{{a}^{\mathrm{2}} {b}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${A}\left(\mathrm{0},{b}\right)\:\:\:{B}\left({a},\mathrm{0}\right)\:\:\:{O}\left(\mathrm{0},\mathrm{0}\right)\:\:\:{and}\:\boldsymbol{{C}}\left(\frac{{ab}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} },\frac{{a}^{\mathrm{2}} {b}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right) \\ $$$${let}\:{C}\:{devides}\:{AB}\:\:\:\:\frac{{m}}{{n}}\:{ratio} \\ $$$$\frac{{ab}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\frac{{m}×\mathrm{0}+{n}×{a}}{{m}+{n}} \\ $$$${mab}^{\mathrm{2}} +{nab}^{\mathrm{2}} ={na}^{\mathrm{3}} +{nab}^{\mathrm{2}} \\ $$$$\frac{{m}}{{n}}=\frac{{a}^{\mathrm{3}} }{{ab}^{\mathrm{2}} }=\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} } \\ $$$${now}\:{slope}\:{of}\:{AB}\:{is}\:{m}_{\mathrm{1}} =\frac{−{b}}{{a}} \\ $$$${let}\:{AB}\:{makes}\:{angle}\:\theta\:{with}\:{OB} \\ $$$${so}\:\:{slope}\left({m}_{\mathrm{1}} \right)=\frac{−{b}}{{a}}={tan}\left(\mathrm{180}^{{o}} −\mathrm{60}^{{o}} \right)=−\sqrt{\mathrm{3}}\: \\ $$$$\frac{{b}}{{a}}=\sqrt{\mathrm{3}}\: \\ $$$$\frac{{m}}{{n}}=\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{3}}\:\:{proved} \\ $$$$\blacksquare\blacksquare\:{if}\:{we}\:{take}\:\theta=\mathrm{30}^{{o}} \\ $$$${slope}=\frac{−{b}}{{a}}={tan}\left(\mathrm{180}^{{o}} −\mathrm{30}^{{o}} \right)=−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}} \\ $$$$\frac{{b}}{{a}}=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}} \\ $$$${then}\:\frac{{m}}{{n}}=\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{3}\: \\ $$$${so}\:{rstion}\:\frac{\mathrm{3}}{\mathrm{1}}\:{also}\:{fulfill}... \\ $$$$ \\ $$$${note}\:{Triangle}\:{OAB} \\ $$$${OA}\bot{OB}\:\:\:{znd}\:{AB}\:{hypotaneous} \\ $$$$ \\ $$
Commented by Kunal12588 last updated on 17/May/19
$${sir}\:{you}\:{still}\:{used}\:{trigonometry}\:\left({tangent}\right) \\ $$
Commented by Kunal12588 last updated on 17/May/19
$${but}\:{i}\:{like}\:{your}\:{way}\:{sir}\:\left({pls}\:{let}\:{it}\:{be}\:{here}\right) \\ $$
Answered by mr W last updated on 17/May/19
Commented by mr W last updated on 17/May/19
$${let}\:{AB}=\mathrm{1},\:{AF}={x} \\ $$$$\Rightarrow{BC}={CE}=\mathrm{1} \\ $$$$\Rightarrow{CD}={x} \\ $$$$\Rightarrow{FE}={ED}=\sqrt{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\Rightarrow{AD}=\mathrm{2}+{x} \\ $$$${AD}^{\mathrm{2}} ={AF}^{\mathrm{2}} +{FD}^{\mathrm{2}} \\ $$$$\Rightarrow\left(\mathrm{2}+{x}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +\left(\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4}+\mathrm{4}{x}+{x}^{\mathrm{2}} ={x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −{x}−\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\left({x}+\mathrm{1}\right)\left({x}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{2} \\ $$$$\Rightarrow{BD}=\mathrm{1}+\mathrm{2}=\mathrm{3} \\ $$$$\Rightarrow{AB}/{BD}=\mathrm{1}/\mathrm{3}\:\Rightarrow{proved} \\ $$
Commented by Kunal12588 last updated on 17/May/19
$${sir}\:{i}\:{tried}\:{pls}\:{check} \\ $$$$\bigtriangleup{AFC}\:{is}\:{equilateral}\: \\ $$$$\Rightarrow{FC}={AC}={AB}+{BC}=\mathrm{2} \\ $$$${FC}={DC} \\ $$$$\Rightarrow{AC}={DC} \\ $$$$\Rightarrow{x}=\mathrm{2} \\ $$$$\frac{{AB}}{{BD}}=\frac{\mathrm{1}}{{x}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Commented by mr W last updated on 17/May/19
$${nice}\:{path}\:{sir}! \\ $$