Monday, June 20, 2011

The Problem of the Week Day 1: Week of 6/19 - 6/25

Since this is the first problem of the week, let me go over how it works. Each week, you have an easy and hard problem to try. The problem is split into five parts, one for each day of the week. If you figure out the answer, contact me and I will tell you if it's right. If it is, I will list you as one of the people who solved this problem. After one month, I will post the answer along with Saturday's post on Cool Math Stuff.

Before you start, don't forget to round all answers you get to the NEAREST TENTH. Tomorrow, you will be solving algebraic equations, and you do not want non-terminating decimals in those!
Easy Problem: When given the two sides of a right triangle, you can use the Pythagorean Theorem to figure out the third. Basically, the longest side is titled c and the others are a and b. Then, you plug the sides into the equation a^2 + b^2 = c^2.

In today's triangle, a = 3 and b = 4. The answer for today's part is c.

Hard Problem: If you only are given one side of your right triangle, but another angle, you can use some simple trigonometry to determine the other two sides. Since the side opposite to your angle divided by c = the sine of that angle, and there is a sine function on some calculators, you can figure it out. All you do is type in your angle, hit the "sine" or "sin" button on your calculator, and multiply it by c to give you another side. Then, use the Pythagorean Theorem to figure out the third.

In our triangle, c = 39.4 in and our angle is 52.5°. Try to figure out a and b (with a being the side you use trigonometry to determine and b being the side you use the Pythagorean Theorem to determine. 

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