Question

va rog rezolvati problema

Answer 1

f) O sa ne folosim de relatia lui Silvester de la d
$\vec{H_{1}H_{2}}=\vec{H_{1}O}+\vec{OH_{2}}=\vec{OH_{2}}-\vec{OH_{1}}=\vec{OM}+\vec{OB}+\vec{OC}-\vec{OM}-\vec{OC}-\vec{OA}=\vec{OB}-\vec{OA}=\vec{AO}+\vec{OB}=\vec{AB}$
$\vec{H_{2}H_{3}}=\vec{H_{2}O}+\vec{OH_{3}}=\vec{OH_{3}}-\vec{OH_{2}}=\vec{OM}+\vec{OB}+\vec{OA}-\vec{OM}-\vec{OC}-\vec{OA}=\vec{OB}-\vec{OC}=\vec{CO}+\vec{OB}=\vec{CB}$
$\vec{H_{1}H_{3}}=\vec{H_{1}O}+\vec{OH_{3}}=\vec{OH_{3}}-\vec{OH_{1}}=\vec{OM}+\vec{OB}+\vec{OA}-\vec{OM}-\vec{OC}-\vec{OB}=\vec{OC}-\vec{OA}=\vec{CO}+\vec{OA}=\vec{CA}$
deci rezulta ca $H_{1}H_{2}=AB$ $H_{2}H_{3}=BC$ si $H_{1}H_{3}=AC$ toate laturile sunt congruente atunci si triunghiurile ABC si $H_{1}H_{2}H_{3}$ sunt congruente
Subiectul IV
Ne folosim de relatiile lui Viete si obtinem urmatoarele relatii
$tga+tgb=-p\Rightarrow \frac{\sin{a}}{\cos{a}}+\frac{\sin{b}}{\cos{b}}=\frac{\sin{a}\cos{b}+\sin{b}\cos{a}}{\cos{a}\cos{b}}=\frac{\sin{a+b}}{\cos{a}\cos{b}}=\frac{\sin{(a+b)}}{\cos{a}\cos{b}}=-p$
si de asemenea
$tga*tgb=q\Rightarrow \frac{\sin{a}\sin{b}}{\cos{a}\cos{b}}=q\Rightarrow 1-\frac{\sin{a}\sin{b}}{\cos{a}\cos{b}}=\frac{\cos{a}\cos{b}-\sin{a}\sin{b}}{\cos{a}\cos{b}}=\frac{\cos{(a+b)}}{\cos{a}\cos{b}}=1-q$
Impartim cele doua relatii si obtinem
$\frac{\sin{(a+b)}}{\cos{(a+b)}}=-\frac{p}{1-q}\Rightarrow \cos{(a+b)}=-\frac{1-q}{p}*\sin{(a+b)}$
Inlocuim pe cos(a+b) in ecuatia lui E
$E=\sin^{2}{(a+b)}+p*\frac{1-q}{p}\sin^{2}{(a+b)}+q*\frac{(1-q)^{2}}{p^{2}}*\sin^{2}{(a+b)}=\sin^{2}{(a+b)}(1+1-q+q\frac{(1-q)^{2}}{p^{2}})$
Mai stim ca
$\sin^{2}{(a+b)}+\cos^{2}{(a+b)}=1\Rightarrow \sin^{2}{(a+b)}+\frac{(1-q)^{2}}{p^{2}}\sin^{2}{(a+b)}=1\Rightarrow \sin^{2}{(a+b)}=\frac{p^{2}}{p^{2}+(1-q)^{2}}$
Atunci
$E=\frac{p^{2}}{p^{2}+(1-q)^{2}}*(1+1-q+q\frac{(1-q)^{2}}{p^{2}})$
Ecuatia deja este in p si q.Daca mai vrei, mai poti face simplificari.