Consider a<span> right triangle with </span><span>hypotenuse of 1.
sin x means to divide the side in front of the angle by </span>hypotenuse<span>
cos x means to divide the side beside the angle by hypotenuse
we know that a^2+b^2=c^2 now divide two sides by c^2

c is 1 also a/c means sinx and b/c is cos x so
sin^2(x)+cos^2(x)=1
</span>
so, is a semi-circle, half a circle, recall a circle has a total of 360°, so half of that will be 180°.
the diameter of that circle is 10, so its radius is half that, or 5.
![\bf \textit{arc's length}\\\\ s=\cfrac{\theta \pi r}{180}~~ \begin{cases} r=radius\\ \theta =angle~in\\ \qquad degrees\\[-0.5em] \hrulefill\\ \theta =180\\ r=5 \end{cases}\implies s=\cfrac{(180)(\pi )(5)}{180}\implies s=5\pi \stackrel{\pi =3.14}{\implies s=15.7}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barc%27s%20length%7D%5C%5C%5C%5C%20s%3D%5Ccfrac%7B%5Ctheta%20%5Cpi%20r%7D%7B180%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%20%5Ctheta%20%3Dangle~in%5C%5C%20%5Cqquad%20degrees%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20%5Ctheta%20%3D180%5C%5C%20r%3D5%20%5Cend%7Bcases%7D%5Cimplies%20s%3D%5Ccfrac%7B%28180%29%28%5Cpi%20%29%285%29%7D%7B180%7D%5Cimplies%20s%3D5%5Cpi%20%5Cstackrel%7B%5Cpi%20%3D3.14%7D%7B%5Cimplies%20s%3D15.7%7D)
Answer:
The first transit on one side of the trench is 18 ft higher than the other side of the trench. The transit is set at 4.5 above ground, therefore the transit lens is 18' + 4.5' = 22.5' above the location on the other side of the trench. The angle of depression is the angle formed by the horizontal and the line of sight to the other side and it is given as 31°. This forms a right angle triangle with height of 22.5' and an interior angle of 31°, Then
tan( 31°) = Height / base, with the base representing the width of the trench
base = Height / tan(31°) = 22.5' / tan(31°) = 37.45' is the width of the trench
Answer:
23,24,26,27 sorry if you get it wrong