Answer :
The question is missing the figure. So, it is in the atachment.
Answer: MN = x[tex]\sqrt{2}[/tex] LN = [tex]\frac{x}{2}.(\sqrt{2} + \sqrt{6} )[/tex]
Step-by-step explanation: The first figure in the attachment is the figure of the question. The second figure is a way to respond this question by tracing the altitude from M to LN as suggested. When an altitude is drawn, it forms a 90° angle with the base, as shown in the drawing. To determine the other angle, you have to remember that all internal angles of a triangle sums up to 180°.
For the triangle on the left of the altitude:
45+90+angle=180
angle = 45
For the triangle on the right:
30+90+angle=180
angle = 60
With the angles, use the Law of Sines, which is relates sides and angles, as follows:
[tex]\frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC}[/tex]
For MN:
[tex]\frac{x}{sin(30)} = \frac{MN}{sin(45)}[/tex]
MN = [tex]\frac{x.sen(45)}{sen(30)}[/tex]
MN = x[tex]\sqrt{2}[/tex]
For LN:
[tex]\frac{LN}{sen(105)} =\frac{x}{sin(30)}[/tex]
LN = [tex]\frac{x.sin(105)}{sin(30)}[/tex]
We can determine sin (105) as:
sin(105) = sin(45+60)
sin(105) = sin(45)cos(60) + cos(45)sin(60)
sin(105) = [tex]\frac{\sqrt{2} }{2}.\frac{1}{2} + \frac{\sqrt{2} }{2}.\frac{\sqrt{3} }{2}[/tex]
sin(105) = [tex]\frac{\sqrt{2} }{4} + \frac{\sqrt{6} }{4}[/tex]
LN = [tex]\frac{x.sin(105)}{sin(30)}[/tex]
LN = [tex]x.(\frac{\sqrt{2} }{4} + \frac{\sqrt{6} }{4} ) .2[/tex]
LN = [tex]\frac{x}{2}.(\sqrt{2} + \sqrt{6} )[/tex]
The expressions for:
MN = x[tex]\sqrt{2}[/tex]
LN = [tex]\frac{x}{2}.(\sqrt{2} + \sqrt{6} )[/tex]
