Could someone help me with question 14?

Pls see attached file.

Results 1 to 4 of 4

- Feb 13th 2011, 09:02 PM #1

- Feb 13th 2011, 10:28 PM #2
Many approaches are possible.

One approach is (wlog) to let the square's sidelength be equal to 1. Then note that $\displaystyle \displaystyle MN = \frac{1}{\sqrt{2}}$ and $\displaystyle \displaystyle MA = \frac{\sqrt{5}}{2}$, use the cosine rule on triangle MAN to get $\displaystyle \cos(\theta)$ and then use the Pythagorean Identity to get $\displaystyle \sin(\theta)$.

(But the problem is that we don't know your mathematical background, so who knows what approaches you will understand and what ones you won't ....)

- Feb 13th 2011, 11:59 PM #3

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Another approach.

Join AC. Now the angle CAN = θ/2. Let the side length of the square be 2. Them DN = 1 and AN = $\displaystyle \sqrt{5}$

Now θ/2 = (45 - x) where angle NAD = x.

Find sin(θ/2) using sin(θ/2) = sin(45 - x). Once you know sin(θ/2) you can find cos(θ/2) then sin(θ) = 2sin(θ/2)cos(θ/2)

- Feb 14th 2011, 07:00 PM #4
I got the answer. Thanks.

The law of cosines is awesome.

I think you meant ma = root 5/2

c squared = a squared + b squared - 2 ab cos theta

1/2 = 5/4 + 5/4 - 5/2 cos theta

cos theta = 4/5

16/25 + sin squared theta = 1

sin theta = 3/5

With my scary persistence mathematical background is really not a problem. If someone gives me an approach, I will eventually understand it.( thanks to google) Sometimes it takes days even weeks though...