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How do I find the measure of the indicated angle, to the nearest tenth of a degree using sine law and cosine law?

Mathematics
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|dw:1337042419972:dw|
Uh, what?
|dw:1337043771132:dw|

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Other answers:

four degrees?
no, it's 104 degrees and the side length of BC is 45 m
|dw:1337046316750:dw|
Let's say that side AB is c and use the law of cosines.
|dw:1337046422125:dw| I found the side length of AC using cosine law: AC^2 = BC^2 + AB^2 - 2(BC)(AB)cosB = 45^2 + 20^2 - 2(45)(20)cos104 = 2860.459412 AC = 53.48 m
\[c^2=20^2+45^2-2(20)(45)\cos 104\]
and then I used cosine law again to find the side length of DC: DC^2 = BC^2 _ BD^2 -2(BC)(BD)cosB = 45^2 + 30^2 -2(45)(30)cos(104 degrees) = 3578.19 I used the triangle BCD
and now I am stuck..
|dw:1337046557115:dw|
Now find the meassure of angle ACB using the law of sines.
|dw:1337046713091:dw|
but what angle am I suppose to find for triangle ABC, when angle B=104 degrees? wait I am confused..
\[\frac{\sin 104}{53.48}=\frac{\sin ACB}{20}\]
Angle ACB = 21.3 degrees
how did you get 21.3?
and the angle of C has to be 7.3 degrees..
|dw:1337046878417:dw|
20(sin 104)/52.48 Then hit the inverse sin button.
how did you get 54.7 and 21.3? :O i'm so sorry
Look above. It tells you how to get the 21.3. Then use the fact that the sum of the angles of a triangle is 180 to get 54.7
ohh yup. But how am I going to find angle C? D:
But you want the angle theta so try this: \[(CD)^2=30^2+45^2-2(30)(45)\cos 104\]
|dw:1337047600966:dw|
i'm sorry but it just says, 'math processing error..'
|dw:1337047693118:dw|
Now use the law of sines to find angle DCB. Subtract 21.3 from DCB and that will be theta.
okay. I'm still kind of confused but thank you so much!
\[\frac{\sin 104}{59.8}=\frac{\sin DCB}{30}\]
What are you confused about?
Mertsj, how come the answer at the back fo my book is 7.8 degrees? o.o
ohh, never mind :$ THANK YOU!!

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