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Roman55 [17]
4 years ago
14

1. WHAT DOES IT MEAN TO SOLVE A RIGHT TRIANGLE?

Mathematics
1 answer:
slega [8]4 years ago
4 0
Hello there!

1) Solving a right triangle means finding the values of the unknown {most likely a variable} 
2) The picture is for the Similarity in Right Triangles, I couldn't explain it.
     How you can use Pythagorean Theorem: The formula for that is (a^2<span> + b^</span>2<span> = c^</span>2<span>). The longest side of a triangle is called the hypotenuse, and the other sides are called legs. You usually use Pythagorean Theorem when you are trying to find an angle of a side. A and B in the formula are the legs and C is the hypotenuse, it makes sense because the hypotenuse is the longest side of the triangle, which means that it has a greater angle than the legs. So say you have a triangle with a leg of 5, a leg of 7, and the hypotenuse is unknown. (If you're a visual person, I also drew it on paint and put it in :P) The formula is a^2 + b^2 = c^2. The ^2 is an exponent, which means that we have to multiply the base by itself that many times. For example, if we had 3^4, we'd have to multiply 3   4 times. (3*3*3*3=81) So 5*5=25 and 7*7=49. It would be better it fill in the formula as you go. So now you have (25+49=c^2). You wouldn't multiply c^2, because we don't know the base yet, it's an unknown, so we keep it at that for a while. So now we just continue with the problem which is basically 25+49 which is 74. (74=c^2) But that's not the answer. We have to find the square root of 74 in order to get the answer because we have to multiply with an exponent, the opposite is square rooting which is what we have to do. (((what sucks is that I reached the max of files to enter for you so your just gonna have to read the rest. sorry))) So the square root of 74 is 8.6. And that's your answer. The legs are 5 and 7, and your hypotenuse is 8.6. But what if you need to find one of the legs. The formula would be a= </span>\sqrt{c^2-b ^2}  ( I hope the insert thing works if it doesn't I said a= square root of c^2-b^2)
I hope you can solve that because I have to go to a testing Class Connect, and I don't think I've learned the other questions yet. I'm sorry, and I hope I've helped!!

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