Please help me find the other sides of the shape?

Question

OABC is a field. A is 88 metres due North of O. B is 146 metres from O on a bearing of 040degrees. C is equidistant from A and from B. The bearing of C from O is 098degrees.
Please help me find the measures of OC ad OB.

Note. by my calculations I found the following:
AB= 96.8m using the cosine law for given two sides and an included angle.
But i could find OC and BC.

Answer 1

hmmm... okay...

I agree with: d(AB) = 96.82831

actually using sine law:
sin(40°)/96 = sin(OAB)/146

this time i do not agree with the approach that OB bisects
I will use analytic geometry to solve this question...
[this is actually ironduke's approach too... but i dont know about his point B]

again set O=(0,0)
then A = (0,88)
B = (146cos(50°) , 146 sin(50°))
= (93.84699,111.84249)

and by distance formula this time... it concurs with your conclusion that d(AB) = 96.82831

In order to locate C, we intersect the perpendicular bisector of AB with the line OC.

to get perpendicular bisector:
midpoint(46.92350, 99.92125)
slope of AB = 0.25406
slope of normal = -3.93612

perpendicular bisector of AB:
y - 99.92125 = -3.93612(x - 46.92350)

now, to get the line OC...
we have bearing of C is 98°
the slope of C is tan(-8°) = -0.14054
thus line OC:
y = -0.14054x

solving simultaneously for
y - 99.92125 = -3.93612(x - 46.92350)
y = -0.14054x

-0.14054x - 99.92125 = -3.93612(x - 46.92350)
(3.93612-0.14054)x = 99.92125 + 3.93612*46.92350
x = 74.98664
y = -10.53862
C = (74.98664,-10.53862)

thus OC = 75.72357
AC = BC = 123.8259

slope of CB = 6.48880
angle of that is ... arctan(6.48880) = 81.23898°

thus
while
it should be clear now that
§

Answer 2

Draw a set of coordinate axes. Make Point O(0,0). Then
Point A is (0,88). Now draw a line AB that is 40 degrees East of North and has a length of 146 meters.. This makes the angle OAB = 140 degrees and not 104 degrees as you have stated. The coordinates of the Point B will be
(146sin40, 146cos40 + 88) = B(93.85, 199.84). Now draw the perpendicular bisector of AB and let it intersect the line drawn from O in direction 8 degrees South of East at Point C.
This means the angle AOC = 98 degrees.

The midpoint of AB is E(93.85/2, (199.84+88)/2)
= E(46.93, 143.92). The slope of AB is 1.19 so the slope of the perpendicular bisector is -.84 . So you can find the equation of the perpendicular bisector and hence find the coordinates of the Point C. The slope of the line OC is -.14.

You now have the field completely defined and should be able to compute OC and BC.