Question Number 153757 by liberty last updated on 10/Sep/21 | ||
Commented by mr W last updated on 11/Sep/21 | ||
$${S}_{\mathrm{1}} ={S}_{\mathrm{2}} =\frac{\mathrm{85}}{\mathrm{4}} \\ $$ | ||
Answered by mr W last updated on 11/Sep/21 | ||
Commented by mr W last updated on 11/Sep/21 | ||
$$\frac{\mathrm{sin}\:\left(\mathrm{45}+\alpha\right)}{\mathrm{sin}\:\alpha}=\frac{\mathrm{10}}{{a}} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}+\mathrm{1}\right)=\frac{\mathrm{10}}{{a}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}=\frac{\mathrm{10}\sqrt{\mathrm{2}}}{{a}}−\mathrm{1} \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{45}+\beta\right)}{\mathrm{sin}\:\beta}=\frac{\mathrm{10}}{{b}} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\frac{\mathrm{1}}{\mathrm{tan}\:\beta}+\mathrm{1}\right)=\frac{\mathrm{10}}{{b}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{tan}\:\beta}=\frac{\mathrm{10}\sqrt{\mathrm{2}}}{{b}}−\mathrm{1} \\ $$$$\alpha+\beta=\mathrm{45}° \\ $$$$\mathrm{tan}\:\left(\alpha+\beta\right)=\mathrm{1} \\ $$$$\frac{\mathrm{tan}\:\alpha+\mathrm{tan}\:\beta}{\mathrm{1}−\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta}=\mathrm{1} \\ $$$$\mathrm{tan}\:\alpha+\mathrm{tan}\:\beta=\mathrm{1}−\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta \\ $$$$\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}+\frac{\mathrm{1}}{\mathrm{tan}\:\beta}=\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}×\frac{\mathrm{1}}{\mathrm{tan}\:\beta}−\mathrm{1} \\ $$$$\frac{\mathrm{10}\sqrt{\mathrm{2}}}{{a}}−\mathrm{1}+\frac{\mathrm{10}\sqrt{\mathrm{2}}}{{b}}−\mathrm{1}=\left(\frac{\mathrm{10}\sqrt{\mathrm{2}}}{{a}}−\mathrm{1}\right)×\left(\frac{\mathrm{10}\sqrt{\mathrm{2}}}{{b}}−\mathrm{1}\right)−\mathrm{1} \\ $$$$\frac{\mathrm{10}\sqrt{\mathrm{2}}}{{a}}+\frac{\mathrm{10}\sqrt{\mathrm{2}}}{{b}}−\mathrm{1}=\frac{\mathrm{100}}{{ab}} \\ $$$$\frac{\mathrm{10}\sqrt{\mathrm{2}}\left({a}+{b}\right)}{{ab}}−\mathrm{1}=\frac{\mathrm{100}}{{ab}} \\ $$$${ab}=\mathrm{15} \\ $$$$\Rightarrow{a}+{b}=\frac{\mathrm{23}\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$${a},{b}\:{are}\:{roots}\:{of} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{23}\sqrt{\mathrm{2}}}{\mathrm{4}}{x}+\mathrm{15}=\mathrm{0} \\ $$$$\Rightarrow{a},{b}=\frac{\left(\mathrm{23}\pm\mathrm{7}\right)\sqrt{\mathrm{2}}}{\mathrm{8}}=\begin{cases}{\mathrm{2}\sqrt{\mathrm{2}}}\\{\frac{\mathrm{15}\sqrt{\mathrm{2}}}{\mathrm{4}}}\end{cases} \\ $$$$\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}=\frac{\mathrm{10}\sqrt{\mathrm{2}}}{\mathrm{2}\sqrt{\mathrm{2}}}−\mathrm{1}=\mathrm{4}\:\Rightarrow\mathrm{tan}\:\alpha=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\frac{\mathrm{1}}{\mathrm{tan}\:\beta}=\frac{\mathrm{10}\sqrt{\mathrm{2}}}{\frac{\mathrm{15}\sqrt{\mathrm{2}}}{\mathrm{4}}}=\frac{\mathrm{5}}{\:\mathrm{3}}\:\Rightarrow\mathrm{tan}\:\beta=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\frac{{c}}{{a}}=\frac{\mathrm{sin}\:\mathrm{45}}{\mathrm{sin}\:\alpha}\:\Rightarrow{c}=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{17}}}}=\mathrm{2}\sqrt{\mathrm{17}} \\ $$$$\frac{{d}}{{b}}=\frac{\mathrm{sin}\:\mathrm{45}}{\mathrm{sin}\:\beta}\:\Rightarrow{d}=\frac{\mathrm{15}\sqrt{\mathrm{2}}}{\:\mathrm{4}\sqrt{\mathrm{2}}×\frac{\mathrm{3}}{\:\sqrt{\mathrm{34}}}}=\frac{\mathrm{5}\sqrt{\mathrm{34}}}{\mathrm{4}} \\ $$$${S}_{\mathrm{1}} =\frac{{cd}\:\mathrm{sin}\:\mathrm{45}°}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}×\mathrm{2}\sqrt{\mathrm{17}}×\frac{\mathrm{5}\sqrt{\mathrm{34}}}{\mathrm{4}}=\frac{\mathrm{85}}{\mathrm{4}} \\ $$$${AE}=\frac{\mathrm{10}}{\mathrm{cos}\:\beta}=\frac{\mathrm{10}}{\frac{\mathrm{5}}{\:\sqrt{\mathrm{34}}}}=\mathrm{2}\sqrt{\mathrm{34}} \\ $$$${AF}=\frac{\mathrm{10}}{\mathrm{cos}\:\alpha}=\frac{\mathrm{10}}{\:\frac{\mathrm{4}}{\:\sqrt{\mathrm{17}}}}=\frac{\mathrm{5}\sqrt{\mathrm{17}}}{\mathrm{2}} \\ $$$${S}_{\mathrm{1}} +{S}_{\mathrm{2}} =\frac{{AE}×{AF}×\mathrm{sin}\:\mathrm{45}°}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}×\mathrm{2}\sqrt{\mathrm{34}}×\frac{\mathrm{5}\sqrt{\mathrm{17}}}{\:\mathrm{2}}=\frac{\mathrm{85}}{\mathrm{2}} \\ $$$$\Rightarrow{S}_{\mathrm{2}} =\frac{\mathrm{85}}{\mathrm{2}}−\frac{\mathrm{85}}{\mathrm{4}}=\frac{\mathrm{85}}{\mathrm{4}}={S}_{\mathrm{1}} \\ $$ | ||
Commented by Ari last updated on 11/Sep/21 | ||
Mr.W. always, the ratio S2/S1 eshte 1 if dhe angle 0f triangle is 45 grade? | ||
Commented by mr W last updated on 11/Sep/21 | ||
$${yes},\:{S}_{\mathrm{1}} ={S}_{\mathrm{2}} \:{is}\:{always}\:{true}. \\ $$ | ||