Answer:
<D + <F = [tex]100^{o}[/tex]
Step-by-step explanation:
Given: <D + <E = [tex]110^{o}[/tex]
<E + <F = [tex]150^{o}[/tex]
But, the sum of angles in a triangle = [tex]180^{o}[/tex].
Thus;
<D + <E + <F = [tex]180^{o}[/tex]
[tex]110^{o}[/tex] + <F = [tex]180^{o}[/tex]
⇒ <F = [tex]180^{o}[/tex] - [tex]110^{o}[/tex]
= [tex]70^{o}[/tex]
<F = [tex]70^{o}[/tex]
Also,
<E + <F = [tex]150^{o}[/tex]
<E + [tex]70^{o}[/tex] = [tex]150^{o}[/tex]
<E = [tex]150^{o}[/tex] - [tex]70^{o}[/tex]
= [tex]80^{o}[/tex]
<E = [tex]80^{o}[/tex]
So that;
<D + <E = [tex]110^{o}[/tex]
<D + [tex]80^{o}[/tex] = [tex]110^{o}[/tex]
<D = [tex]110^{o}[/tex] - [tex]80^{o}[/tex]
= [tex]30^{o}[/tex]
<D = [tex]30^{o}[/tex]
⇒ <D = [tex]30^{o}[/tex], <E = [tex]80^{o}[/tex], and <F = [tex]70^{o}[/tex].
Therefore;
<D + <F = [tex]30^{o}[/tex] + [tex]70^{o}[/tex]
= [tex]100^{o}[/tex]
