Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the correct position in the answer box. Release your mouse button when the item is place. If you change your mind, drag the item to the trashcan. Click the trashcan to clear all your answers. Using the technique in the model above, find the missing sides in this 30°-60°-90° right triangle. Short = 2 Long =
@mathstudent55
So far what I have is that the hypotenuse is 4 and the long side is 6, correct?
No. You are correct that the hypotenuse is 4. The short leg is 2, and the hypotenuse is 2 times the short leg, so the hypotenuse is 2 * 2 = 4. The long leg is not 3 times the short leg. That would make the long leg longer than the hypotenuse. The hypotenuse is always the longest side of a right triangle. The long leg is \(\sqrt 3\) times the short leg. The long leg is \(\sqrt 3\) times 2, so the long leg is \(2 \sqrt 3\)
Ohh. so it's 3.4?
\(\sqrt 3\) is approximately 1.7, so \(2\sqrt 3 \approx 3.4\) (\(\approx\) means "is approximately equal to"
Alright.
Since the long leg is approx. 3.4, you have a short leg of 2, and long leg of 3.4, and a hypotenuse of 4. As you can see, the lengths of the sides are in the correct order of sizes: short leg, long leg, hypotenuse: 2, 3.4, 4
Okay, so what's the equation I'd use for it?
You use the ratio of the lengths of the sides of a 30-60-90 triangle: \(1 : \sqrt 3 : 2\)
In this problem, you could not start by using the Pythagoras theorem because you only knew one side. You had to use the ratios.
I know that. now there is a thing at the bottom of the problem where you plug in the question for how you got the answer and then the answer
Remember that the hypotenuse is always the longest side of a right triangle. If you calculate side lengths, and you get a leg longer that the hypotenuse, you know you made a mistake.
Yeah, that's kind of obvious. Alright, well thanks for helping with this one.
Short leg = 2. Hypotenuse = 2 * short leg = 2 * 2 = 4 Now you have two choices for the long leg: 1) use ratios long leg = \(\sqrt 3\) * short leg = \(\sqrt 3 \times 2 = 2 \sqrt 3\) 2) Pythagoras theorem \(a^2 + b^2 = c^2\) \(2^2 + b^2 = 4^2\) \(4 + b^2 = 16\) \(b^2 = 12\) \(b = \sqrt {12} = \sqrt {4 \times 3} = 2 \sqrt 3\)
You wrote "Yeah, that's kind of obvious. " that the hypotenuse is always the longest side, but earlier, you seemed to have forgotten it when you knew the hypotenuse was 4 and you asked if the long leg was 6.
Yeah, Sometimes i get confused lol.
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