Find: LBC [ft] A Ax,y= ( [ft], [ft])

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Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Cx,y= (34.10 [ft], 94.53 [ft]) o N5 10’E Ox,y= (0 [ft], 0 [ft]) R=100.5 [ft] C B y 124 128 132 136 R Find the arc length from Point B to Point C, in feet. [pause] In this problem, Line segment A B ---- R North x O

Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Cx,y= (34.10 [ft], 94.53 [ft]) o N5 10’E Ox,y= (0 [ft], 0 [ft]) R=100.5 [ft] C B y 124 128 132 136 R and Arc B C are connected at point B. R North x O

Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Cx,y= (34.10 [ft], 94.53 [ft]) o N5 10’E Ox,y= (0 [ft], 0 [ft]) R=100.5 [ft] C B y 124 128 132 136 R The coordinates of points A, ---- R North x O

Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Cx,y= (34.10 [ft], 94.53 [ft]) o N5 10’E Ox,y= (0 [ft], 0 [ft]) R=100.5 [ft] C B y 124 128 132 136 R C and O are provided, where point O is at the origin of the ---- R North x O

Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Cx,y= (34.10 [ft], 94.53 [ft]) o N5 10’E Ox,y= (0 [ft], 0 [ft]) R=100.5 [ft] C B y 124 128 132 136 R cartesian coordinate system, and the north arrow point upwards. Also, the radius of the curve is given, ---- R North x O

Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Cx,y= (34.10 [ft], 94.53 [ft]) o N5 10’E Ox,y= (0 [ft], 0 [ft]) R=100.5 [ft] C B y 124 128 132 136 R as well as the bearing of line segment A B. Also note that line segment A B is not tangent ---- R North x O

Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Cx,y= (34.10 [ft], 94.53 [ft]) o N5 10’E Ox,y= (0 [ft], 0 [ft]) R=100.5 [ft] C B y 124 128 132 136 R to the curve. [pause] To find the length of arc B C, ---- R x O

Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Cx,y= (34.10 [ft], 94.53 [ft]) o N5 10’E Ox,y= (0 [ft], 0 [ft]) R=100.5 [ft] C B LBC = R * I R we can multiply the radius of the curve, R, by the interior angle, ---- I O

Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Cx,y= (34.10 [ft], 94.53 [ft]) o N5 10’E Ox,y= (0 [ft], 0 [ft]) R=100.5 [ft] C B LBC = R * I R I. The problem statement provides the radius, which is --- I interior radius angle O

Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Cx,y= (34.10 [ft], 94.53 [ft]) o N5 10’E Ox,y= (0 [ft], 0 [ft]) R=100.5 [ft] C B LBC = R * I R 100.5 feet, but we still don’t know the interior angle. In the figure, we’ll define point O prime, as the northward projection ---- I interior radius angle O

Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Cx,y= (34.10 [ft], 94.53 [ft]) o N5 10’E Ox,y= (0 [ft], 0 [ft]) O’ R=100.5 [ft] C B LBC = R * I R of point O onto arc B C. That way, we can equate interior Angle I to Angle ----- I interior radius angle O

Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Cx,y= (34.10 [ft], 94.53 [ft]) o N5 10’E Ox,y= (0 [ft], 0 [ft]) O’ R=100.5 [ft] C B LBC = R * I B O O prime plus Angle O prime O C, which we’ll call --- I interior radius angle I=ABOO’+AO’OC O

Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Cx,y= (34.10 [ft], 94.53 [ft]) o N5 10’E Ox,y= (0 [ft], 0 [ft]) O’ R=100.5 [ft] C B I1 I2 LBC = R * I Angle I 1 plus Angle I 2. [pause] We can determine Angle I 2, using trigonometry and by knowing ---- I interior radius angle I=ABOO’+AO’OC O I=I1+I2

Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Cx,y= (34.10 [ft], 94.53 [ft]) o N5 10’E Ox,y= (0 [ft], 0 [ft]) O’ C I=I1+I2 B Cx-Ox I1 I2 I2=tan-1 Cy-Oy the coordinates of Points O and C. After plugging in ---- I R O

Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Cx,y= (34.10 [ft], 94.53 [ft]) o N5 10’E Ox,y= (0 [ft], 0 [ft]) O’ C I=I1+I2 B Cx-Ox I1 I2 I2=tan-1 Cy-Oy the appropriate values, angle I 2 equals, --- I R O

Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Cx,y= (34.10 [ft], 94.53 [ft]) o N5 10’E Ox,y= (0 [ft], 0 [ft]) O’ C I=I1+I2 B Cx-Ox I1 I2 I2=tan-1 Cy-Oy 19.84 degrees. [pause] Similarly, we can solve for angle I 1 using ---- I R o I2=19.84 O

Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Cx,y= (34.10 [ft], 94.53 [ft]) o N5 10’E Ox,y= (0 [ft], 0 [ft]) O’ C I=I1+I2 B Cx-Ox I1 I2 I2=tan-1 Cy-Oy the coordinates of points B and O. We already know the the coordinates of point O, --- I R o I2=19.84 Ox-Bx O I1=tan-1 By-Oy

? ? Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Cx,y= (34.10 [ft], 94.53 [ft]) o N5 10’E Ox,y= (0 [ft], 0 [ft]) O’ C I=I1+I2 B Cx-Ox I1 I2 I2=tan-1 Cy-Oy but we don’t know the coordinates of point B. We can determine the coordinates of point B since we know ---- I R ? ? o I2=19.84 Ox-Bx O I1=tan-1 By-Oy

Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Ox,y= (0 [ft], 0 [ft]) -Bx o N5 10’E I1=tan-1 By O’ C B I1 I2 it is a point on the line A B, and also a point on the circular arc, B C. The general equation for a line ---- I R O

Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Ox,y= (0 [ft], 0 [ft]) -Bx o y N5 10’E I1=tan-1 By O’ C y - yo = m (x - xo) B I1 I2 equation of line in 2 dimensions is, y minus y not equals, m, times the quantity, x minus x not. In this equation, --- I R in 2-dimensions x O

Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Ox,y= (0 [ft], 0 [ft]) -Bx o y equation N5 10’E I1=tan-1 of line By O’ C y - yo = m (x - xo) B I1 I2 known y known x y not and x not, are known y and x coordinates for a point on the line, and variable m ---- I R coordinate coordinate x O

Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Ox,y= (0 [ft], 0 [ft]) -Bx o y equation N5 10’E I1=tan-1 of line By O’ C y - yo = m (x - xo) B I1 I2 known y known x is the slope of the line, in y over x. [pause] Since we know ---- I R coordinate coordinate x slope O

Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Ox,y= (0 [ft], 0 [ft]) -Bx o y equation N5 10’E I1=tan-1 of line By O’ C y - yo = m (x - xo) B I1 I2 known y known x Point A is on line A B, then we can substitute negative 62.71 and 247.65 in for ---- I R coordinate coordinate x slope O

Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Ox,y= (0 [ft], 0 [ft]) -Bx o y equation N5 10’E I1=tan-1 of line By O’ C y - yo = m (x - xo) B I1 I2 known y known x x not and y not, respectively. Also, we know the slope of line A B is the tangent of the quantity --- I R coordinate coordinate x slope O

Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) N Ox,y= (0 [ft], 0 [ft]) -Bx o y equation N5 10’E I1=tan-1 of line By O’ C y - yo = m (x - xo) B I1 I2 known y known x 90 degrees minus the northeast bearing of line segment A B, ---- I R coordinate coordinate x o slope=tan(90-AABN) O

Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) N Ox,y= (0 [ft], 0 [ft]) -Bx o y equation N5 10’E I1=tan-1 of line By O’ C y - yo = m (x - xo) B I1 I2 known y known x So, 90 minus 5 and 1/6th equals, 84.83, and the tangent of 84.83 degrees equals, ---- I R coordinate coordinate x o slope=tan(90-AABN) O o =tan(84.83 )

Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) N Ox,y= (0 [ft], 0 [ft]) -Bx o y equation N5 10’E I1=tan-1 of line By O’ C y - yo = m (x - xo) B I1 I2 known y known x 11.05. [pause] Now we have our equation of line A B, --- I R coordinate coordinate x o slope=tan(90-AABN) O o =tan(84.83 ) =11.05

Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) N Ox,y= (0 [ft], 0 [ft]) o y equation N5 10’E slope=11.05 of line O’ C y - yo = m (x - xo) B I1 I2 known y known x which simplifies to, y equals --- I R coordinate coordinate x o slope=tan(90-AABN) O o =tan(84.83 ) =11.05

Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Ox,y= (0 [ft], 0 [ft]) o y equation N5 10’E slope=11.05 of line O’ C y - yo = m (x - xo) B I1 I2 y=11.05*x+940.6 [ft] 11.05 times x, plus 940.6 feet. [pause] Next we’ll solve for the equation of the circle ---- I R Line AB x O

Find: LBC [ft] R=100.5 [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Ox,y= (0 [ft], 0 [ft]) o y equation N5 10’E slope=11.05 of line O’ C y - yo = m (x - xo) B I1 I2 y=11.05*x+940.6 [ft] which contains circular arc, B C. To do so, we’ll plug in --- I R (x-xo)2+(y-yo)2=R2 x O equation of circle

Find: LBC [ft] R=100.5 [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Ox,y= (0 [ft], 0 [ft]) o y equation N5 10’E slope=11.05 of line O’ C y - yo = m (x - xo) B I1 I2 y=11.05*x+940.6 [ft] the x and y coordinates of the center of the circle for x not and y not, --- I R (x-xo)2+(y-yo)2=R2 x O equation center of circle coordinates

Find: LBC [ft] R=100.5 [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Ox,y= (0 [ft], 0 [ft]) o y equation N5 10’E slope=11.05 of line O’ C y - yo = m (x - xo) B I1 I2 y=11.05*x+940.6 [ft] as well as the radius, R. After plugging in the appropriate values, --- I R (x-xo)2+(y-yo)2=R2 x O radius equation center of circle coordinates

Find: LBC [ft] R=100.5 [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Ox,y= (0 [ft], 0 [ft]) o y equation N5 10’E slope=11.05 of line O’ C y - yo = m (x - xo) B I1 I2 y=11.05*x+940.6 [ft] the equation of the circle simplifies to --- I R (x-xo)2+(y-yo)2=R2 x O radius equation center of circle coordinates

Find: LBC [ft] R=100.5 [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Ox,y= (0 [ft], 0 [ft]) o y equation N5 10’E slope=11.05 of line O’ C y - yo = m (x - xo) B I1 I2 y=11.05*x+940.6 [ft] x squared plus y squared equals 10,100, feet squared. [pause] At this point we have ---- I R (x-xo)2+(y-yo)2=R2 x O equation x2+y2=10,100 [ft2] of circle

Find: LBC [ft] R=100.5 [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Ox,y= (0 [ft], 0 [ft]) o y N5 10’E y=11.05*x+940.6 [ft] line O’ x2+y2=10,100 [ft2] B I1 I2 circle 2 equations, and 2 unknown variables. So we can solve for x by substituting in ---- I R x O

Find: LBC [ft] R=100.5 [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Ox,y= (0 [ft], 0 [ft]) o y N5 10’E y=11.05*x+940.6 [ft] line O’ x2+y2=10,100 [ft2] B I1 I2 circle 11.05 times x plus 940.6 feet for y, and after some algebra, we get x squared plus ---- I R x O

Find: LBC [ft] R=100.5 [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Ox,y= (0 [ft], 0 [ft]) o N5 10’E y=11.05*x+940.6 [ft] line O’ x2+y2=10,100 [ft2] B circle x2 + 168.9 * x [ft] + 7,105 [ft2] = 0 168.9 feet times x, plus 7,105 feet squared equals 0. Now we can solve for x using ---

Find: LBC [ft] R=100.5 [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Ox,y= (0 [ft], 0 [ft]) o N5 10’E y=11.05*x+940.6 [ft] line O’ x2+y2=10,100 [ft2] B circle x2 + 168.9 * x [ft] + 7,105 [ft2] = 0 the quadratic equation, where coefficient a, b and c are 1, --- -b± b2-4 * a * c x= 2 * a

Find: LBC [ft] R=100.5 [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Ox,y= (0 [ft], 0 [ft]) o N5 10’E y=11.05*x+940.6 [ft] line O’ x2+y2=10,100 [ft2] B b c a circle 1x2 + 168.9 * x [ft] + 7,105 [ft2] = 0 168.9, and 7,105, respectively. After---- -b± b2-4 * a * c x= 2 * a

Find: LBC [ft] R=100.5 [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Ox,y= (0 [ft], 0 [ft]) o N5 10’E y=11.05*x+940.6 [ft] line O’ x2+y2=10,100 [ft2] B b c a circle 1x2 + 168.9 * x [ft] + 7,105 [ft2] = 0 plugging these values in, x equals ---- -b± b2-4 * a * c x= 2 * a

Find: LBC [ft] R=100.5 [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Ox,y= (0 [ft], 0 [ft]) o N5 10’E y=11.05*x+940.6 [ft] line O’ C x2+y2=10,100 [ft2] B b c a circle 1x2 + 168.9 * x [ft] + 7,105 [ft2] = 0 negative 89.63 feet and negative 79.27 feet. These 2 x values represent the 2 x coordinates corresponding to the intersection --- -b± b2-4 * a * c x= 2 * a x=-89.63 [ft],-79.27 [ft]

Find: LBC [ft] R=100.5 [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) Ox,y= (0 [ft], 0 [ft]) o N5 10’E y=11.05*x+940.6 [ft] line O’ C x2+y2=10,100 [ft2] B b a circle c 1x2 + 168.9 * x [ft] + 7,105 [ft2] = 0 of line A B, and the circle which contains arc B C. From our coordinate system, --- -b± b2-4 * a * c x= 2 * a x=-89.63 [ft],-79.27 [ft]

Find: LBC [ft] x R=100.5 [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) o Ox,y= (0 [ft], 0 [ft]) N5 10’E y=11.05*x+940.6 [ft] line O’ C x2+y2=10,100 [ft2] B b a circle c 1x2 + 168.9 * x [ft] + 7,105 [ft2] = 0 x increases as we move from left to right, therefore, --- x -b± b2-4 * a * c x= 2 * a x=-89.63 [ft],-79.27 [ft]

Find: LBC [ft] x R=100.5 [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) o Ox,y= (0 [ft], 0 [ft]) N5 10’E y=11.05*x+940.6 [ft] line O’ C x2+y2=10,100 [ft2] B b a circle c 1x2 + 168.9 * x [ft] + 7,105 [ft2] = 0 the larger value of x represents to x coordinate of point B. Substituting negative 79.27 feet in for --- x -b± b2-4 * a * c x= 2 * a x=-89.63 [ft],-79.27 [ft]

Find: LBC [ft] R=100.5 [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) o Ox,y= (0 [ft], 0 [ft]) N5 10’E y=11.05*x+940.6 [ft] line O’ C x2+y2=10,100 [ft2] B b a circle c 1x2 + 168.9 * x [ft] + 7,105 [ft2] = 0 x, the y coordinate of Point B equals, --- -b± b2-4 * a * c x= 2 * a x=-89.63 [ft],-79.27 [ft]

Find: LBC [ft] R=100.5 [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) o Ox,y= (0 [ft], 0 [ft]) N5 10’E y=11.05*x+940.6 [ft] line O’ C x2+y2=10,100 [ft2] B b a circle c 1x2 + 168.9 * x [ft] + 7,105 [ft2] = 0 64.67 feet. [pause] By knowing the coordinates of point B, --- y=64.67 [ft] -b± b2-4 * a * c x= 2 * a x=-89.63 [ft],-79.27 [ft]

Find: LBC [ft] R=100.5 [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) o Ox,y= (0 [ft], 0 [ft]) N5 10’E Bx,y= (-79.27 [ft], 64.67 [ft]) O’ C B b c a 1x2 + 168.9 * x [ft] + 7,105 [ft2] = 0 we can return to our equation for angle I 1, --- y=64.67 [ft] -b± b2-4 * a * c x= 2 * a x=-89.63 [ft],-79.27 [ft]

Find: LBC [ft] R=100.5 [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) o o Ox,y= (0 [ft], 0 [ft]) I2=19.84 N5 10’E Bx,y= (-79.27 [ft], 64.67 [ft]) O’ -Bx C I1=tan-1 By B I1 I2 plug in the x and y coordinates of point B, --- I R y=64.67 [ft] x O x=-79.27 [ft]

Find: LBC [ft] R=100.5 [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) o o Ox,y= (0 [ft], 0 [ft]) I2=19.84 N5 10’E Bx,y= (-79.27 [ft], 64.67 [ft]) O’ -Bx C I1=tan-1 By B I1 I2 and calculate Angle I 1 to be, --- I R x O

Find: LBC [ft] R=100.5 [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) o o Ox,y= (0 [ft], 0 [ft]) I2=19.84 N5 10’E Bx,y= (-79.27 [ft], 64.67 [ft]) O’ -Bx C I1=tan-1 By B I1 I2 o I1=50.79 50.79 degrees. Now we can compute ---- I R x O

Find: LBC [ft] R=100.5 [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) o o Ox,y= (0 [ft], 0 [ft]) I2=19.84 N5 10’E Bx,y= (-79.27 [ft], 64.67 [ft]) O’ -Bx C I1=tan-1 By B I1 I2 o I1=50.79 the value of Angle I, by adding 50.79 degrees and --- I R I = I1 + I2 x O

Find: LBC [ft] R=100.5 [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) o o Ox,y= (0 [ft], 0 [ft]) I2=19.84 N5 10’E Bx,y= (-79.27 [ft], 64.67 [ft]) O’ -Bx C I1=tan-1 By B I1 I2 o I1=50.79 19.84 degrees, which equals 70.63 --- I R I = I1 + I2 x O

Find: LBC [ft] R=100.5 [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) o o Ox,y= (0 [ft], 0 [ft]) I2=19.84 N5 10’E Bx,y= (-79.27 [ft], 64.67 [ft]) O’ -Bx C I1=tan-1 By B I1 I2 o I1=50.79 degrees. [pause] Lastly, the length of arc B C equals --- I R I = I1 + I2 x o O I = 70.63

Find: LBC [ft] R=100.5 [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) o o Ox,y= (0 [ft], 0 [ft]) I2=19.84 N5 10’E Bx,y= (-79.27 [ft], 64.67 [ft]) O’ I = I1 + I2 C o B I = 70.63 I1 I2 the radius of the curve times the interior angle swept out from Point B to Point C. In this equation, --- I R LBC = R * I x O

Find: LBC [ft] R=100.5 [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) o o Ox,y= (0 [ft], 0 [ft]) I2=19.84 N5 10’E Bx,y= (-79.27 [ft], 64.67 [ft]) O’ I = I1 + I2 C o B I = 70.63 I1 I2 angle I should be in units of radians, therefore we’ll multiply ---- I R LBC = R * I x O radians

Find: LBC [ft] π R=100.5 [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) o o Ox,y= (0 [ft], 0 [ft]) I2=19.84 N5 10’E Bx,y= (-79.27 [ft], 64.67 [ft]) O’ I = I1 + I2 C o π B I = 70.63 I1 I2 * 180 70.63 degrees by, PI over 180, which makes I equal to ---- I R LBC = R * I x O radians

Find: LBC [ft] π R=100.5 [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) o o Ox,y= (0 [ft], 0 [ft]) I2=19.84 N5 10’E Bx,y= (-79.27 [ft], 64.67 [ft]) O’ I = I1 + I2 C o π B I = 70.63 I1 I2 * 180 1.233 radians. After plugging in the values of the radius and the --- I = 1.233 [rad] I R LBC = R * I x O radians

Find: LBC [ft] π R=100.5 [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) o o Ox,y= (0 [ft], 0 [ft]) I2=19.84 N5 10’E Bx,y= (-79.27 [ft], 64.67 [ft]) O’ I = I1 + I2 C o π B I = 70.63 I1 I2 * 180 interior angle, the length of arc B C equals, --- I = 1.233 [rad] I R LBC = R * I x O radians

Find: LBC [ft] π R=100.5 [ft] A Ax,y= (-62.71 [ft], 247.65 [ft]) o o Ox,y= (0 [ft], 0 [ft]) I2=19.84 N5 10’E Bx,y= (-79.27 [ft], 64.67 [ft]) O’ I = I1 + I2 C o π B I = 70.63 I1 I2 * 180 123.9 feet. [pause] I = 1.233 [rad] I R LBC = R * I x LBC = 123.9 [ft] O

Find: LBC [ft] π R=100.5 [ft] Ax,y= (-62.71 [ft], 247.65 [ft]) o Ox,y= (0 [ft], 0 [ft]) I2=19.84 Bx,y= (-79.27 [ft], 64.67 [ft]) I = I1 + I2 124 128 132 136 o π I = 70.63 * 180 When looking over the possible solutions, --- I = 1.233 [rad] LBC = R * I LBC = 123.9 [ft]

Find: LBC [ft] π R=100.5 [ft] Ax,y= (-62.71 [ft], 247.65 [ft]) o Ox,y= (0 [ft], 0 [ft]) I2=19.84 Bx,y= (-79.27 [ft], 64.67 [ft]) I = I1 + I2 124 128 132 136 o π I = 70.63 * 180 the answer is A. I = 1.233 [rad] LBC = R * I LBC = 123.9 [ft] AnswerA

? Index σ’v = Σ γ d γT=100 [lb/ft3] +γclay dclay 1 Find: σ’v at d = 30 feet (1+wc)*γw wc+(1/SG) σ’v = Σ γ d d Sand 10 ft γT=100 [lb/ft3] 100 [lb/ft3] 10 [ft] 20 ft Clay = γsand dsand +γclay dclay A W S V [ft3] W [lb] 40 ft text wc = 37% ? Δh 20 [ft] (5 [cm])2 * π/4