SRL Archive - Mathematics

This will probably be a joke to some of you. If I can teach this in my class I will receive 5% up my grade lol.

So basically, try to give me formulas, and the law of sines/cosines you can use, and any SOHCAHTOA that you want. But these I need to be able to explain it( PLUS, these aren't even the real numbers, but they are close – I just can't remember)

1. Find X
2. find the height of that stop sign. It is basically if it were touching the sun.


The up hill to the base of the sign in 215 ft. Ya!

Thanks :D

Obviously wrong. Never mind.

not sure where your getting this 33 from.

But I think I might have figured it out.

The dotted line is not a measurement, the road that is leading up to the sign is 215 ft.

Then, the angle between the road and the dotted line in 20 degrees.
The angle between the road and the line to the SUN is 53 degrees.

Sol'n:

Let y = the distance between the bottom of the post and the lower ground level.
Find y using the sine law and the relation between y, the interior angle = 20, the hill distance and the right angle.
y/sin(20) = 215/sin(90)
y = 215sin(20) ('cause sin(90) = 1)

Let h = the height of the sign.
z = the total height
h = z - y

Let d = the base of the triangles
Find d by using the pythagorean theorem.
d = (215^(2) + y^(2))^(1/2)
d = (215^(2) + (215sin(20))^(2))^(1/2)

Let a = the angle between x and the sign
The sum of all interior angles is equal to 180, so
180 = 90 + 53 + a (rearrange for a)
a = 180 - 53 - 90
a = 37

Find x using sine law:
x/sin(90) = d/sin(37)
x = d/sin(37) ('cause sin(90) = 1)
x = (((215)^(2) + (215sin(20))^(2))^(1/2))/sin(37)

(YAY ONE DOWN)

Find z, use pythagorean's theorem:
z = (x^(2) - d^(2))^(1/2)
z = ((d/sin(37))^(2) - d^(2))^(1/2)
z = ((d^(2) - (sin^(2)(37))(d)^(2))/sin^(2)(37))^(1/2)
z = (d^(2) . (1 + sin^(2)(37))/(sin^(2)(37)))^(1/2) ('cause simplify radical, (d^(2))^(1/2) = +d)
z = (+d)((1 + sin^(2)(37))/(sin^(2)(37)))^(1/2)

Now that we have z, lets find h!
h = z - y
h = (+d)((1 + sin^(2)(37))/(sin^(2)(37)))^(1/2) - (215sin(20)) ('cause substituting everything back in..)
h = (((215)^(2) + (215sin(20))^(2))^(1/2)) . ((1 + sin^(2)(37))/(sin^(2)(37)))^(1/2) - (215sin(20)) ('cause that ^)

There, its in exact form even..

That took WAYYY too long to type out.

Also, no calculator used, and I *probably* could simplify h a bit but I wont :p

Hey thanks. That is definaetly what I needed, and I'm glad someone cuold understand the drawing!! Thanks both of you!

Although it would have been easier with a calculator, such as one provided on most computers :duh:

Draw Better Next Time Please! It would help a lot!

shutup, yohojo. It was in sub 1 minute, :p

BUT OK, JUST FOR YOU!!!

From: Ejjman1
Hey thanks. That is definaetly what I needed, and I'm glad someone cuold understand the drawing!! Thanks both of you!

Although it would have been easier with a calculator, such as one provided on most computers :duh:

I'm not allowed a calculator in either of my math courses..

I'm used to it :)

TriBase = 215Cos20;
x = TriBase / Cos53;
stopsign^2 = 215^2 + x^2 - 215*x*cos(53 - 20 = 33).
stopsign = sqrt(215^2 + x^2 - 215*x*cos33)
(Cosine rule, yay).

Edit: Double checked and mine produces the same result as n3s's for the stopsign height :)

From: mixster
TriBase = 215Cos20;
x = TriBase / Cos53;
stopsign^2 = 215^2 + x^2 - 215*x*cos(53 - 20 = 33).
stopsign = sqrt(215^2 + x^2 - 215*x*cos33)
(Cosine rule, yay).

Edit: Double checked and mine produces the same result as n3s's for the stopsign height :)

I think we just proved the law of cosines.. :<