How do I find the sides of a right triangle using sine, cos, or tan, with only the three angles and the hypotenuse
1 answer:
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Beforehand let me apologize for my sloppy handwriting. I'm a lefty, so please deal with me. . Anyway, the equation you listed didn't really have a x, because a linear equation is y= mx+b, and anything with x is the slope. But I did see y and a number on one side, so I'm like maybe I can get these two alone. So here's what I did:
2y + 4=0
-4 -4 I subtracted 4 on both sides. Why because you would want "y" alone.
Next,
2y= -4
Now to get "y" alone you want to divide on both sides.
2y= -4/ 2
y= -2
Now you're probably thinking "how do you graph it? There's no "x" in the equation." Well, you just graph it. Since the answer is y= -2, you go to the y axis, look for -2, and place a line to indicate that is the equation. And to make it clear, remember y is the y-intercept.
I really do hope this helps you, if not message me. I'll be happy to help, and again I'm sorry for my handwriting!
2y + 4=0
-4 -4 I subtracted 4 on both sides. Why because you would want "y" alone.
Next,
2y= -4
Now to get "y" alone you want to divide on both sides.
2y= -4/ 2
y= -2
Now you're probably thinking "how do you graph it? There's no "x" in the equation." Well, you just graph it. Since the answer is y= -2, you go to the y axis, look for -2, and place a line to indicate that is the equation. And to make it clear, remember y is the y-intercept.
I really do hope this helps you, if not message me. I'll be happy to help, and again I'm sorry for my handwriting!