Attn. Physics People

[sailor]

Now that I look at it again I realize it has to be a joke as nothing makes any sense.

[Manlob]

Maybe I was joking about the complex number solution, but I think this method is easier than yours. There is no need to find t and the trig can be done mentally since most people know cos(60 degrees)=0.5, while your method requires computation of arctan(51.96/30).

Step by step (using same triangle as you did):

sin(x)=15/t
cos(x)=150/t[sup]2[/sup]

Square sin(x) and use the trig identity sin[sup]2[/sup]x=1-cos[sup]2[/sup]x, so 1-cos[sup]2[/sup]=225/t[sup]2[/sup].

You can either eliminate t[sup]2[/sup], or just note that multiplying cos(x)=150/t[sup]2[/sup] by 1.5 gives 225/t[sup]2[/sup], which equals 1-cos[sup]2[/sup]x. Either way you are left with a quadratic which equals zero and factors into (0.5-cos(x)) and (2+cos(x)). So cos(x) must be either 0.5 or -2.

The original post said tan(x)=1.5/sin(x), and if you use tan=sin/cos this can also be converted into the same quadradic for cos(x).

[hightechburrito]

I couldn’t have done it the easy way could I? I just had to do it the fun way!

At least I got it right.

[sailor]

Manlob, I resolve the three sides of the triangle without using any trig functions as 30, sqrt(300) and 60. To define the angle I chose the ATAN function because it is the one commonly found in calculators which may not have ACOS() but you can do ACOS just as well as you have all three sides.

My resolution is based on the direct analysis of the problem as stated. I still cannot see where you get directly from the information stated that sin(x) = 15/t etc. I am not saying it is mistaken. I am saying it does not follow immediately from the information given. (Or maybe I am missing something).

Can you explain how you get 15/t and 150/t^2 ?? I can’t even see the dimension correctly. These two magnitudes have to resolve as expressed in units of distance (m). I can’t see where you get the numbers or the units. Can you explain it?

[AWB]

*Originally posted by sailor *
**AWB, Rectangular means in general anything with a right angle. Rect means right, angular, angle. Rectangular triangle is a triangle with a 90 degree angle.

In common English usage you were thinking of rectangle but even this is imprecise and the strictly correct way of expressing what you have in mind would be “rectangular parallelepiped”.**

I’ll concede this one. My “Mathematics Dictionary” doesn’t ever refer to right triagles as rectangular; that’s where I was coming from.

But then looking at its definition of a rectangle (“A parallelogram with one angle a right angle and therefore all of its angles right angles” [emphasis mine]), I see that only one angle being 90[sup]o[/sup] is the basis of its definition.

Looking further, the Lycos dictionary said “having one or more right angles”.

I’ve never heard the term before, which is why I thought it was a good joke. I’ve always heard “right triangle”.

[sailor]

Right Triangle (maybe as opposed to the wrong triangle is common usage, but us nerds like to talk with precission because if not, some other doper will come out of the woodwork to point out how wrong everything was said.

[Saltire]

]

*Originally posted by sailor *
**AWB, Rectangular means in general anything with a right angle. Rect means right, angular, angle. Rectangular triangle is a triangle with a 90 degree angle.

In common English usage you were thinking of rectangle but even this is imprecise and the strictly correct way of expressing what you have in mind would be “rectangular parallelepiped”.**

*Originally posted by sailor *
**Right Triangle (maybe as opposed to the wrong triangle is common usage, but us nerds like to talk with precission because if not, some other doper will come out of the woodwork to point out how wrong everything was said. **

For instance, when they tell you that a parallelepiped is a solid polyhedron, while the discussion has been two-dimensional. What you meant to say is “rectangular parallelogram,” I believe.

Normally, I avoid correcting such misnomers, but I just couldn’t resist after you invited me out of the woodwork:D

[sailor]

>> What you meant to say is “rectangular parallelogram,” I believe

Yep, what on earth was I thinking?

[Manlob]

sailor says:

Manlob, I resolve the three sides of the triangle without using any trig functions as 30, sqrt(300) and 60.

The question didn’t ask for the sides of the triangle or the time, it asked for the angle, and you did use a trig function to find that.

Can you explain how you get 15/t and 150/t^2 ?? I can’t even see the dimension correctly. These two magnitudes have to resolve as expressed in units of distance (m). I can’t see where you get the numbers or the units. Can you explain it?

It follows from the triangle: Length of side opposite angle x: 3t. Length of hypotenuse: t[sup]2[/sup]/5. sin(x)= (opposite)/(hypotenuse)=(3t)/(t[sup]2[/sup]/5)=15/t. If you are worried about units: sin(x)= (15 sec)/t; cos(x)=(150 sec[sup]2[/sup])/t[sup]2[/sup].

I don’t have much of a problem with your method of solving this, just that the question asks for the angle, and you solve for the time, then the lengths of the sides, then the angle. I prefer to do algebra instead of arithmetic so I set up an equation that can be solved for the angle.

To avoid solving for unecessary information or using any trig functions, start with the right triangle: hypotenuse length of t[sup]2[/sup]/5, and sides of length 30 and 3t.
30[sup]2[/sup] + (3t)[sup]2[/sup] = (t[sup]2[/sup]/5)[sup]2[/sup]
Rearrange and factor:
0= (t[sup]2[/sup]/5)[sup]2[/sup] -45(t[sup]2[/sup]/5)-900 = (t[sup]2[/sup]/5-60)(t[sup]2[/sup]/5+15)
So t[sup]2[/sup]/5=60=Length of hypotenuse
Hypotenuse is twice the length of the side adjacent to angle x, so x=60 degrees.