Consideram triunghiul DEF ascutiunghic si isoscel, DE = DF.
[tex]\mathcal{A} =\dfrac{DE\cdot DF\cdot \sin D}{2} = \dfrac{4\sqrt6\cdot 4\sqrt6\ \cdot \sin D}{2} = 48 \cdot \sin D\ \ \ \ (1)
[/tex]
[tex]Dar,\ \mathcal{A} = 24\ \ \ (2)
\\\;\\
(1),\ (2) \Rightarrow 48\cdot \sin D =24 \Rightarrow \sin D = \dfrac{24}{48} =\dfrac{1}{2} \Rightarrow m(\hat{D}) =30^0
[/tex]
[tex]\Delta {DFE} -isoscel\ \Rightarrow m(\hat{F}) = m(\hat{E}) = \dfrac{180^0-30^0}{2}= 75^0[/tex]
