Expert: Sherman D. - 10/13/2004

Question
To the nearest degree, the measure of the acute angle formed by the x-axis and the line that contains points A(4,7) and B(8,10) is
1. 53.1
2. 36.9
3. 48.6
4. 41.4
The answer is 36.9, but I can't find a way to do it. Please help.

Answer
(4,7) and (8,10)

m = (10 - 7)/(8 - 4)
m = 3/4

(4,7), m = (3/4)

7 = (3/4)(4) + b
7 = 3 + b
b = 4

y = (3/4)x + 4

0 = (3/4)x + 4
-4 = (3/4)x
-16 = 3x

x = (-16/3) or -5 1/3

using this, you get points at

(8,10), (8,0), and ((-16/3),0)

to go from (8,0) to ((-16/3),0) you get (40/3)
to go from (8,10) to ((-16/3),0), you get

d = sqrt(((-16/3) - 8)^2 + (0 - 10)^2)
d = sqrt((-40/3)^2 + (-10)^2)
d = sqrt((1600/9) + 100)
d = sqrt(-700/9)
d = sqrt(2500/9)
d = 50/3

you end up with
Hypothenuse (50/3)
Adjacent Side = 40/3, bottom side

Cos X = (40/3)/(50/3)
Cos X = (40/3) * (3/50)
Cos X = (40/50)
Cos X = (4/5)

Inverse cosine both sides and you get

X = 36.8698976584

or best rounded to

x = 36.9

So your answer is 2.) 36.9

at one time i had 0 = (3/4)x + 4, the reason for this is, because the problem said that the angle formed on the x-axis, so the y had to equal 0, so that gave me x was -16/3

so going from there to where the (8,10) met up with the x-axis at 8, so to go from 8 to -16/3 gave me a distance of (40/3) and using the furtherest point which was (8,10) and to go from there to ((-16/3),0) gave me a distance of (50/3) using the distance formula. I couldn't just say that the distance for (8,10) to ((-16/3),0) was also (40/3) since this is a diagonal, and if you know the pythagorean thereom then you know that when dealing with a right triangle, the diagonal is always longer than either side.

Info found at http://wright.nasa.gov/airplane/trig.html and http://www.explorelearning.com/index.cfm?method=cResource.dspDetail&ResourceID=6...

I hope you understand what i did.