Question

What are expressions for MN and LN? Hint Construct the altitude from M to LN.

Answer 1

The question is missing the figure. So, it is in the atachment.

Answer: MN = x$\sqrt{2}$  LN = $\frac{x}{2}.(\sqrt{2} + \sqrt{6} )$

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:

$\frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC}$

For MN:

$\frac{x}{sin(30)} = \frac{MN}{sin(45)}$

MN = $\frac{x.sen(45)}{sen(30)}$

MN = x$\sqrt{2}$

For LN:

$\frac{LN}{sen(105)} =\frac{x}{sin(30)}$

LN = $\frac{x.sin(105)}{sin(30)}$

We can determine sin (105) as:

sin(105) = sin(45+60)

sin(105) = sin(45)cos(60) + cos(45)sin(60)

sin(105) = $\frac{\sqrt{2} }{2}.\frac{1}{2} + \frac{\sqrt{2} }{2}.\frac{\sqrt{3} }{2}$

sin(105) = $\frac{\sqrt{2} }{4} + \frac{\sqrt{6} }{4}$

LN = $\frac{x.sin(105)}{sin(30)}$

LN = $x.(\frac{\sqrt{2} }{4} + \frac{\sqrt{6} }{4} ) .2$

LN = $\frac{x}{2}.(\sqrt{2} + \sqrt{6} )$

The expressions for:

MN = x$\sqrt{2}$

LN = $\frac{x}{2}.(\sqrt{2} + \sqrt{6} )$