The given triangle is a right-angled triangle
[tex]FE=12\text{units and m}\angle F=35^o\text{.}[/tex][tex]\text{Here m}\angle E=90^o[/tex]Using the triangle sum property, we get
[tex]m\angle D+m\angle F+m\angle E=180^o_{}[/tex][tex]\text{ Substitute }\angle mF=35^o\text{ and m}\angle E=90^o,\text{ we get}[/tex][tex]m\angle D+35^o+90^o=180^o[/tex][tex]m\angle D+125^o=180^o[/tex]Adding -125 degrees on both sides, we get
[tex]m\angle D+125^o-125^o=180^o-125^o[/tex][tex]m\angle D=55^o[/tex]Recall the formula for cosine
[tex]\text{cos}\theta=\frac{Adjacent\text{ side}}{\text{Hypotenuse}}[/tex]Substitute adjacent side =FE=12 units and Hypotenuse =FD and angle is 35 degree.
[tex]\cos 35^o=\frac{12}{FD}[/tex]Taking reciprocal and multiplying 12 on both sides.
[tex]\frac{12}{\cos 35^o}=\frac{FD}{12}\times12[/tex][tex]FD=\frac{12}{\cos35^o}[/tex][tex]\text{Use }\cos 35^o=0.819.[/tex][tex]FD=\frac{12}{0.819}=14.65\text{ units }[/tex]Recall the formula sine.
[tex]\sin \theta=\frac{opposite\text{ side}}{\text{hypotenuse}}[/tex]Substitute opposide side=DE and Hypotenuse =FD=14.65 units and angle is 35 degrees.
[tex]\sin 35^o=\frac{DE}{14.65}[/tex][tex]Use\sin 35^o=0.573[/tex][tex]0.573=\frac{DE}{14.65}[/tex][tex]DE=0.573\times14.65=8.39=8.40[/tex]Hence the required answer is
[tex]m\angle D=55^o\text{ , DE=8.40 units and DF=14.65 units.}[/tex]Hence the option B is correct.
