Answer:
[tex]MN=5\sin 15^{\circ}\approx 1.29\ m[/tex]
Step-by-step explanation:
Since MN ⊥ ED, then triangle MND is right triangle. In this triangle,
- MD = 5 m, because point M is the midpoint of the segment AD;
- ∠MDN=15° (given).
Thus,
[tex]\sin \angle MND=\dfrac{MN}{MD},\\ \\\sin 15^{\circ}=\dfrac{MN}{5},\\ \\MN=5\sin 15^{\circ}\approx 1.29\ m.[/tex]
