1. ## Sin problem

Could someone help me with question 14?
Pls see attached file.

2. Originally Posted by Veronica1999
Could someone help me with question 14?
Pls see attached file.
Many approaches are possible.

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

(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 ....)

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

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...