D,E,F are the feet of the altitudes in triangle ABC. Find angle DEF (in degrees).
+26367 D,E,F are the feet of the altitudes in triangle ABC.
Find angle DEF (in degrees).
\(\text{Let $\angle[DEF]=x$} \\ \text{Let $AB=a$, $AC=b$, $BC=c$} \\ \text{Let $\angle[CBE]=90^\circ-65^\circ=25^\circ$} \\ \text{Let $\angle[EBF]=90^\circ-75^\circ=15^\circ$} \\ \text{Let $BF=c*\cos(40^\circ)$, $BD=a*\cos(40^\circ)$} \\ \text{Let $BE=\dfrac{a*c}{b}\sin(40^\circ)$ } \\ \text{Let $\angle[BDE]=D$, $\angle[BFE]=E$} \\ \text{Let $x =360^\circ-(D+E+40^\circ)=320^\circ-(D+E)$}\)
\(\begin{array}{|rcll|} \hline \tan(D) &=& \dfrac{BE*\sin(25^\circ)}{BD-BE*\cos(25^\circ)} \\\\ \tan(D) &=& \dfrac{\dfrac{a*c}{b}*\sin(40^\circ)*\sin(25^\circ)}{a*\cos(40^\circ)-\dfrac{a*c}{b}\sin(40^\circ)*\cos(25^\circ)} \times \dfrac{a}{a} \\\\ \tan(D) &=& \dfrac{\dfrac{c}{b}*\sin(40^\circ)*\sin(25^\circ)}{\cos(40^\circ)-\dfrac{c}{b}\sin(40^\circ)*\cos(25^\circ)} \\\\ && \boxed{ \dfrac{c}{b} = \dfrac{\sin(75^\circ)} {\sin(40^\circ)} } \\\\ \tan(D) &=& \dfrac{\sin(75^\circ)*\sin(25^\circ)}{\cos(40^\circ)-\sin(75^\circ)*\cos(25^\circ)} \\\\ D &=& -75^\circ~ \text{or}~ \mathbf{D=105^\circ} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \tan(E) &=& \dfrac{BE*\sin(15^\circ)}{BF-BE*\cos(15^\circ)} \\\\ \tan(E) &=& \dfrac{\dfrac{a*c}{b}*\sin(40^\circ)*\sin(15^\circ)}{c*\cos(40^\circ)-\dfrac{a*c}{b}\sin(40^\circ)*\cos(15^\circ)} \times \dfrac{c}{c}\\\\ \tan(E) &=& \dfrac{\dfrac{a}{b}*\sin(40^\circ)*\sin(15^\circ)}{\cos(40^\circ)-\dfrac{a}{b}\sin(40^\circ)*\cos(15^\circ)} \\\\ && \boxed{ \dfrac{a}{b} = \dfrac{\sin(65^\circ)} {\sin(40^\circ)} } \\\\ \tan(E) &=& \dfrac{\sin(65^\circ)*\sin(15^\circ)}{\cos(40^\circ)-\sin(65^\circ)*\cos(15^\circ)} \\\\ E &=& -65^\circ~ \text{or}~ \mathbf{E=115^\circ} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline x &=& 320^\circ-(D+E) \\ x &=& 320^\circ-(105^\circ+115^\circ) \\ x &=& 320^\circ-220^\circ \\ \mathbf{x} &=& \mathbf{100^\circ} \\ \hline \end{array}\)
