2.
We can find angle BAC as follows
6^2 = 2(5)^2 - 2(5^2) cos(BAC)
[36 - 50] / (-50) = cos(BAC)
-14/-50 = cos(BAC)
7/25 = cos (BAC)
arccos ( 7/25) = BAC
And since BAC is an inscribed angle, the BOC has twice its measure = 2arccos(7/25)
And cos [ 2 arccos (7/25) ] = - 527/ 625
And OC = OB
So......using the Law of Cosines
6^2 = 2(OC)^2 - 2(OC)^2 cos [2arccos(7/25) ]
36 = 2(OC)^2 - 2(OC)^2 [ -527/ 625]
36 = 2(OC)^2 + 2(OC)^2 [ 527/625]
18 = OC^2 [ 1 + 527/625]
18 = OC^2 [ 1152/ 625]
18 * 625 / 1152 = OC^2 = 11250/ 1152 = 625/ 64
So.....the area of triangle OBC =
(1/2) OC^2 * sin (BOC)
(1/2) (625/64) * sin [ 2 arccos(7/25) ] =
(625/ 128) * sin [ 2 arccos (7/25) ] =
(625 / 128) * (336/625) =
336/128 =
( 21 / 8 ) units^2
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