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please i need help before tomorrow
where's the "figures" ???
in the attachment below
do you see it
ok.. i see it.... that arrow way on the right is pointing north?
to find angle B, notice that the sum of all the angles in any quadrilateral is 360 degrees. so add those known angles then subtract the sum from 360 degrees ...
since the angles are given in degree-minute-seconds, give the measure of angle B the same way.
is angle b 360
no... angle B = 360 - (angle A + angle D + angle C)
what else do i need to do? for the other angles
If i'm not wrong, bearing of AB will be 84*21'
you're wrong though....too bad
\[ \large \angle B=121º29' \]
hold on a sec.
I think it will be: 96º11' -84º21' = 11º50 Then: 180 - (11º50 +89º38) to get the bearing of AB
what about BC and DA?
the bearing of BC is N42º57'W
bearing of DA N11º50'W
bearing of AB E11º28'N
i chose you as best answer cause you stayed to help:)
u r welcome. and thank you the key is to draw a coordinate system on each point.
oh okay, yeah i didn't do that guess i should have
this theorem of geometry is key |dw:1350960824103:dw|
oh wait so angle be is 121.484 correct
i posted the answer for that (my first post)
guess i didn't know because everyone was confused
ok. good luck.