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

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milliex51Best ResponseYou've already chosen the best response.0
dw:1337042419972:dw
 one year ago

jiberjiberBest ResponseYou've already chosen the best response.0
dw:1337043771132:dw
 one year ago

milliex51Best ResponseYou've already chosen the best response.0
no, it's 104 degrees and the side length of BC is 45 m
 one year ago

MertsjBest ResponseYou've already chosen the best response.1
Let's say that side AB is c and use the law of cosines.
 one year ago

milliex51Best ResponseYou've already chosen the best response.0
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
 one year ago

MertsjBest ResponseYou've already chosen the best response.1
\[c^2=20^2+45^22(20)(45)\cos 104\]
 one year ago

milliex51Best ResponseYou've already chosen the best response.0
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
 one year ago

milliex51Best ResponseYou've already chosen the best response.0
and now I am stuck..
 one year ago

MertsjBest ResponseYou've already chosen the best response.1
Now find the meassure of angle ACB using the law of sines.
 one year ago

milliex51Best ResponseYou've already chosen the best response.0
but what angle am I suppose to find for triangle ABC, when angle B=104 degrees? wait I am confused..
 one year ago

MertsjBest ResponseYou've already chosen the best response.1
\[\frac{\sin 104}{53.48}=\frac{\sin ACB}{20}\]
 one year ago

MertsjBest ResponseYou've already chosen the best response.1
Angle ACB = 21.3 degrees
 one year ago

milliex51Best ResponseYou've already chosen the best response.0
how did you get 21.3?
 one year ago

milliex51Best ResponseYou've already chosen the best response.0
and the angle of C has to be 7.3 degrees..
 one year ago

MertsjBest ResponseYou've already chosen the best response.1
20(sin 104)/52.48 Then hit the inverse sin button.
 one year ago

milliex51Best ResponseYou've already chosen the best response.0
how did you get 54.7 and 21.3? :O i'm so sorry
 one year ago

MertsjBest ResponseYou've already chosen the best response.1
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
 one year ago

milliex51Best ResponseYou've already chosen the best response.0
ohh yup. But how am I going to find angle C? D:
 one year ago

MertsjBest ResponseYou've already chosen the best response.1
But you want the angle theta so try this: \[(CD)^2=30^2+45^22(30)(45)\cos 104\]
 one year ago

milliex51Best ResponseYou've already chosen the best response.0
i'm sorry but it just says, 'math processing error..'
 one year ago

MertsjBest ResponseYou've already chosen the best response.1
Now use the law of sines to find angle DCB. Subtract 21.3 from DCB and that will be theta.
 one year ago

milliex51Best ResponseYou've already chosen the best response.0
okay. I'm still kind of confused but thank you so much!
 one year ago

MertsjBest ResponseYou've already chosen the best response.1
\[\frac{\sin 104}{59.8}=\frac{\sin DCB}{30}\]
 one year ago

MertsjBest ResponseYou've already chosen the best response.1
What are you confused about?
 one year ago

milliex51Best ResponseYou've already chosen the best response.0
Mertsj, how come the answer at the back fo my book is 7.8 degrees? o.o
 one year ago

milliex51Best ResponseYou've already chosen the best response.0
ohh, never mind :$ THANK YOU!!
 one year ago
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