1. Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft])
Cx,y= (34.10 [ft], [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

2. Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft])
Cx,y= (34.10 [ft], [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

3. Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft])
Cx,y= (34.10 [ft], [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

4. Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft])
Cx,y= (34.10 [ft], [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

5. Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft])
Cx,y= (34.10 [ft], [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

6. Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft])
Cx,y= (34.10 [ft], [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

7. Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft])
Cx,y= (34.10 [ft], [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

8. Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft])
Cx,y= (34.10 [ft], [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

9. Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft])
Cx,y= (34.10 [ft], [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

10. Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft])
Cx,y= (34.10 [ft], [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

11. Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft])
Cx,y= (34.10 [ft], [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

12. Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft])
Cx,y= (34.10 [ft], [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

13. Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft])
Cx,y= (34.10 [ft], [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

14. Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft])
Cx,y= (34.10 [ft], [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

15. Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft])
Cx,y= (34.10 [ft], [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

16. Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft])
Cx,y= (34.10 [ft], [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

17. Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft])
Cx,y= (34.10 [ft], [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

18. ? ? Find: LBC [ft] A Ax,y= (-62.71 [ft], 247.65 [ft])
Cx,y= (34.10 [ft], [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

19. 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

20. 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

21. 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

22. 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

23. 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 and in for ---- I R
coordinate
coordinate x
slope O

24. 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

25. 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

26. 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 degrees equals, ---- I R
coordinate
coordinate x o
slope=tan(90-AABN) O o
=tan(84.83 )

27. 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
[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

28. 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

29. 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 [ft]
11.05 times x, plus feet. [pause] Next we’ll solve for the equation of the circle ---- I R
Line AB x O

30. 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 [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

31. 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 [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

32. 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 [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

33. 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 [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

34. 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 [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

35. 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 [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

36. 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 [ft]
line O’
x2+y2=10,100 [ft2] B I1 I2
circle
11.05 times x plus feet for y, and after some algebra, we get x squared plus ---- I R x O

37. 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 [ft]
line O’
x2+y2=10,100 [ft2] B
circle
x * 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 ---

38. 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 [ft]
line O’
x2+y2=10,100 [ft2] B
circle
x * 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

39. 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 [ft]
line O’
x2+y2=10,100 [ft2] B b c a
circle
1x * x [ft] + 7,105 [ft2] = 0
168.9, and 7,105, respectively. After----
-b± b2-4 * a * c x=
2 * a

40. 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 [ft]
line O’
x2+y2=10,100 [ft2] B b c a
circle
1x * x [ft] + 7,105 [ft2] = 0
plugging these values in, x equals ----
-b± b2-4 * a * c x=
2 * a

41. 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 [ft]
line O’ C
x2+y2=10,100 [ft2] B b c a
circle
1x * x [ft] + 7,105 [ft2] = 0
negative feet and negative feet. These 2 x values represent the 2 x coordinates corresponding to the intersection ---
-b± b2-4 * a * c x=
2 * a
x= [ft], [ft]

42. 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 [ft]
line O’ C
x2+y2=10,100 [ft2] B b a
circle c
1x * 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= [ft], [ft]

43. 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 [ft]
line O’ C
x2+y2=10,100 [ft2] B b a
circle c
1x * 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= [ft], [ft]

44. 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 [ft]
line O’ C
x2+y2=10,100 [ft2] B b a
circle c
1x * x [ft] + 7,105 [ft2] = 0
the larger value of x represents to x coordinate of point B. Substituting negative feet in for --- x
-b± b2-4 * a * c x=
2 * a
x= [ft], [ft]

45. 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 [ft]
line O’ C
x2+y2=10,100 [ft2] B b a
circle c
1x * x [ft] + 7,105 [ft2] = 0
x, the y coordinate of Point B equals, ---
-b± b2-4 * a * c x=
2 * a
x= [ft], [ft]

46. 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 [ft]
line O’ C
x2+y2=10,100 [ft2] B b a
circle c
1x * 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= [ft], [ft]

47. 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= ( [ft], [ft]) O’ C B b c a
1x * 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= [ft], [ft]

48. 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= ( [ft], [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= [ft]

49. 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= ( [ft], [ft]) O’
-Bx C
I1=tan-1 By B I1 I2
and calculate Angle I 1 to be, --- I R x O

50. 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= ( [ft], [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

51. 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= ( [ft], [ft]) O’
-Bx C
I1=tan-1 By B I1 I2 o
I1=50.79
the value of Angle I, by adding degrees and --- I R
I = I1 + I2 x O

52. 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= ( [ft], [ft]) O’
-Bx C
I1=tan-1 By B I1 I2 o
I1=50.79
19.84 degrees, which equals I R
I = I1 + I2 x O

53. 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= ( [ft], [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

54. 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= ( [ft], [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

55. 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= ( [ft], [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

56. 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= ( [ft], [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

57. 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= ( [ft], [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 = [rad] I R
LBC = R * I x O
radians

58. 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= ( [ft], [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 = [rad] I R
LBC = R * I x O
radians

59. 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= ( [ft], [ft]) O’
I = I1 + I2 C o π B
I = 70.63 I1 I2 *
180
123.9 feet. [pause]
I = [rad] I R
LBC = R * I x
LBC = [ft] O

60. 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= ( [ft], [ft])
I = I1 + I2
124
128
132
136 o π
I = 70.63 *
180
When looking over the possible solutions, ---
I = [rad]
LBC = R * I
LBC = [ft]

61. 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= ( [ft], [ft])
I = I1 + I2
124
128
132
136 o π
I = 70.63 *
180
the answer is A.
I = [rad]
LBC = R * I
LBC = [ft]
AnswerA

62. ? 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