Answer:
s[tex]\sqrt{5}[/tex]
Step-by-step explanation:
Hi there! I've graphed a square, with its midpoints so that you can visualize it better (below).
1) Starting with a square ABCD, a square since it has four equal angles, its sides (s) they have the same size.
2) The first statement says the sin M∠AN can be expressed in terms of [tex]\frac{m}{n}[/tex] and m, and n are relatively prime positive integers, in other words, there is only one divisor of m and n: 1. And n can never be zero.
With that in mind, let's start calculating:
Sin MAN=
[tex]sin A =\frac{m}{n} \\ m=?\\[/tex]
m=?
Well, m is the hypotenuse of Δ AND
Then, [tex]m^{2} =s^{2}+(\frac{s}{2})^2\\ m=\sqrt{s^{2}+\frac{s^{2}}{4}} \\ m=\sqrt{\frac{5s^{2}}{4}} \\ m=\frac{s\sqrt{5}}{2}[/tex]
n=?
Well, n is also a hypotenuse. But the Δ ABM one.
Then, similarly to m
n²=(s)²+(s²/2)²
n²=s²+s²/4
n²=5s²/4
n=[tex]\frac{s\sqrt{5}}{2}[/tex]
3) Finally, we're now able to find the sum m +n
m + n = ?
[tex]m =(\frac{s\sqrt{5}}{2} )+ n =(\frac{s\sqrt{5}}{2}) =\frac{2s\sqrt{5}}{2} =s\sqrt{5}[/tex]
And review and test our work.
In the first part of the question, it was said the value of the sin of MAN, i.e. the sin of ∠A = [tex]\frac{m}{n}[/tex]. The distance between a midpoint and a vertex is the same. So, MA = NA
You can choose a number, let's say a square whose sides measure 4 units and do the test.
So MA = m and n=NA and sin ∠A is 90º
sin ∠A = MA/NA
sin 90º =MA/NA
1=1 correct
