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Serggg [28]
4 years ago
7

A 180 ft long board is cut into 3 pieces. The second piece of board is twice as long as the first. The third piece of board is s

ix times as long as the first. What is the length of the first piece of the board?
Mathematics
1 answer:
andrew11 [14]4 years ago
8 0
So, let's let w be the length of the first piece of the board.
(I was going to do l (small L) but it's too hard to see on computers)

second piece is 2w since it's twice the length of first.
third piece is 6w, since it's six times longer than the first.
and these three pieces were one one 180ft board, therefore, that means:
w + 2w + 6w = 180

now let's solve for w (length of first piece)!
w + 2w + 6w = 180
7w = 180
w = 180/7 ≈ 25.71ft

Therefore, the first piece of the board would be approximately 25.71ft.

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Answer:  \frac{\sqrt{26}}{26}

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========================================================

\theta is in Q1 so \sin(\theta) > 0

Use the trig identity below and plug in the given value to get...

\sin\left(\frac{x}{2}\right) = \pm \sqrt{\frac{1-\cos(x)}{2}}\\\\\sin\left(\frac{2\theta}{2}\right) = \pm \sqrt{\frac{1-\cos(2\theta)}{2}}\\\\\sin(\theta) = \sqrt{\frac{1-\cos(2\theta)}{2}} \ \ \text{since } \ \sin(\theta) > 0\\\\\sin(\theta) = \sqrt{\frac{1-\frac{12}{13}}{2}}\\\\

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------------------------------

Edit:

To find cos(theta), we use the pythagorean identity below

\sin^2(\theta) + \cos^2(\theta) = 1\\\\\cos^2(\theta) = 1-\sin^2(\theta)\\\\\cos(\theta) = \sqrt{1-\sin^2(\theta)}\ \ \text{cosine is positive in Q1}\\\\\cos(\theta) = \sqrt{1-\left(\frac{1}{\sqrt{26}}\right)^2}\\\\\cos(\theta) = \sqrt{1-\frac{1}{26}}\\\\\cos(\theta) = \sqrt{\frac{25}{26}}\\\\\cos(\theta) = \frac{\sqrt{25}}{\sqrt{26}}\\\\\cos(\theta) = \frac{5}{\sqrt{26}}\\\\\cos(\theta) = \frac{5\sqrt{26}}{\sqrt{26}*\sqrt{26}}\\\\\cos(\theta) = \frac{5\sqrt{26}}{26}\\\\

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3 years ago
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