Using Pythagorean theorem

I’m reviewing my trig before helping my cousin tommorow.

Been awhile. :wink:

This web site is good. Except they leave the exponents on the numbers after doing the math.

That’s not how I remember doing these problems.

Look at example 2. They solve for the length of the base. Square 24 = 576 and 26 squared is 676

Why did they leave the exponents on???

That’s wrong. The operation has been performed. You squared the number.

http://www.mathwarehouse.com/geometry/triangles/how-to-use-the-pythagorean-theorem.php

Has anything changed? I took trig in high school and college. Cal 2 focused on trigonometric equations. It’s been 25 years but I still remember exponents and roots.

I’m not going to say anything if my cousin’s textbook does it this way. I don’t want to confuse her.

But that seems wrong. How can you show an exponent on a number after you already performed the operation?

It’s a typo (someone copied the equation, substituted 676 for 26, and forgot to erase the superscript, maybe) - and a particularly obnoxious one, since it’s going to confuse the people that this page is supposed to help.

They did the same thing in their test question. They make you click a button for the answer. They show exponents after they’ve already done the operation.

I knew that had to be wrong.

I haven’t forgotten that much since college.

I helped my daughters with their high school algebra a few years ago. That made me sweat a little. ;). It’s amazing how quickly we forget this stuff unless we use it.

They did it with both numbers in 2 lines. Yes, it was clearly a typo. 24^2 = 576 and 26^2 = 676

I’m still guessing based on no evidence whatever that it’s a copy-and-paste attempt gone awry…

Isn’t that spelted “cosine”? … it’s been awhile since I’ve taken trig too …

I really enjoyed trig, even the trigonometric functions weren’t bad. I liked it much better than straight Algebra.

Cal 2 (using Calculus on trigonometric functions) was tough. That’s when I discovered I couldn’t earn a minor in math. I took my B in Cal 2 and was thankful for it. That was my last advanced math class.

You missed all the fun stuff by quitting before Differential Equations.

A man has to know his limits. :wink:

Cal 2 was one of the hardest classes I ever took. I’m sure my professor rounded my average up to a B.

Back then the Calculus textbook was huge and very expensive. IIRC it was used for all three semesters. Students always worried the textbook would change before they could finish.

Another problem with that webpage:

They have a very cutesy animated gif that shows c[sup]2[/sup] units of water being poured into the a[sup]2[/sup] and b[sup]2[/sup] areas; then poured back. The problem is that only at the beginning or end of the pouring is the amount of water equal to c[sup]2[/sup] ! In between, it looks like the gif creator just “eyeballed” the water levels and did a poor job.

(The animated gif uses transparencies so most of the 188 frames contain very little information. Is there an easy way, e.g. with Gimp, to collapse the transparencies to get the 188 full frames? I’d be masochistic enough to count the water pixels, but it would be too tedious without this first step.)

I went over triangles with my cousin earlier this afternoon. Calculating the length of the sides, interior angles, and using right triangles to find the area of any triangle. Really glad that I reviewed the material first. :wink:

I enjoyed it. I think she has a better understanding of working with triangles. She’s a smart kid. Hopefully she’ll be ready for her test next week.

The teacher gave her the classic snowball fight problem. I remembered it from high school.

Two boys stand back to back. Each boy holds a snowball. They step forward 4 ft. Turn right and walk 7 ft. They turn towards each other and throw their snowball.

How far apart are they?

I printed off a sheet of graph paper. Had her plot the points and draw the two triangles. Its an easy problem after she saw it drawn.

4^2 + 7^2 = c^2
65 = c^2

each boy is 8.1 ft from the starting point.

They are 16.2 ft apart.

You could ‘amuse’ her with the North Pole trick question.

The two boys are standing at the North Pole. One heads due South for 10 metres and then turns East for another 10 metres. The second boy also heads due South, but after 10 metres he turns due West for the same distance. How far apart are they?

I’m sure there’s a joke here but I’m not getting it.

I’ll get some graph paper and see if I can figure it out.

There’s no way of telling*, though it can be no less than 0 nor more than [st]5*pi[/st] meters.

*If you’re at the north pole, EVERY direction is due south.

edit: actually, no more than 20 meters if they somehow manage to end up directly opposite each other. (I was thinking of the diameter as 10 meters when it’s 20.

Ok. :wink:

I knew there was a joke in there somewhere.

The Pythagorean Theorem is good for only one thing in real life.

You’re making a joke about the fact that “due south” is ambiguous when starting from the north pole. But that has nothing to do with the Pythagorean Theorem. What’s more relevant is the fact that the Pythagorean Theorem does not apply on surfaces which have curvature, for example the surface of the Earth. It only works on flat surfaces.

Consider the following triangle. One vertex is the north pole. From there, follow the 0-degree longitude line all the way to the equator. Now turn left 90 degrees and follow the equator 10,000 km to the 90 degree east longitude line. Turn left 90 degrees and go back to the north pole again. You have a three-sided triangle where each side is 10,000 km long and all three angles measure 90 degrees. Now try to apply the Pythagorean Theorem to this triangle. How can a2 + b2 = c2 when a, b, and c are all equal to each other? Clearly, this triangle is a “right” triangle because it has a 90 degree angle (in fact, it has three of them). So tell me, which side is the hypotenuse?

You could easily construct an example on the surface of the Earth by choosing three cities which are very far apart, chosen carefully so that two of the lines that connect them form a right angle but the other two don’t. Then you’d know which line is the hypotenuse, but you’d still get a wrong answer when you tried to apply the Pythagorean Theorem to find its length.

Hunters walk from their camp 10 miles due South, then 10 miles due East … shoot a bear … and walk 10 miles back to camp … what color was the bear?

White

Math jokes …

Has the site been updated? I don’t see any errors.