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milliex51
 2 years ago
How do I find the measure of the indicated angle, to the nearest tenth of a degree using sine law and cosine law?
milliex51
 2 years ago
How do I find the measure of the indicated angle, to the nearest tenth of a degree using sine law and cosine law?

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milliex51
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1337042419972:dw

jiberjiber
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1337043771132:dw

milliex51
 2 years ago
Best ResponseYou've already chosen the best response.0no, it's 104 degrees and the side length of BC is 45 m

Mertsj
 2 years ago
Best ResponseYou've already chosen the best response.1Let's say that side AB is c and use the law of cosines.

milliex51
 2 years ago
Best ResponseYou've already chosen the best response.0dw: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

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

milliex51
 2 years ago
Best ResponseYou've already chosen the best response.0and 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

Mertsj
 2 years ago
Best ResponseYou've already chosen the best response.1Now find the meassure of angle ACB using the law of sines.

milliex51
 2 years ago
Best ResponseYou've already chosen the best response.0but what angle am I suppose to find for triangle ABC, when angle B=104 degrees? wait I am confused..

Mertsj
 2 years ago
Best ResponseYou've already chosen the best response.1\[\frac{\sin 104}{53.48}=\frac{\sin ACB}{20}\]

Mertsj
 2 years ago
Best ResponseYou've already chosen the best response.1Angle ACB = 21.3 degrees

milliex51
 2 years ago
Best ResponseYou've already chosen the best response.0how did you get 21.3?

milliex51
 2 years ago
Best ResponseYou've already chosen the best response.0and the angle of C has to be 7.3 degrees..

Mertsj
 2 years ago
Best ResponseYou've already chosen the best response.120(sin 104)/52.48 Then hit the inverse sin button.

milliex51
 2 years ago
Best ResponseYou've already chosen the best response.0how did you get 54.7 and 21.3? :O i'm so sorry

Mertsj
 2 years ago
Best ResponseYou've already chosen the best response.1Look 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

milliex51
 2 years ago
Best ResponseYou've already chosen the best response.0ohh yup. But how am I going to find angle C? D:

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

milliex51
 2 years ago
Best ResponseYou've already chosen the best response.0i'm sorry but it just says, 'math processing error..'

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

milliex51
 2 years ago
Best ResponseYou've already chosen the best response.0okay. I'm still kind of confused but thank you so much!

Mertsj
 2 years ago
Best ResponseYou've already chosen the best response.1\[\frac{\sin 104}{59.8}=\frac{\sin DCB}{30}\]

Mertsj
 2 years ago
Best ResponseYou've already chosen the best response.1What are you confused about?

milliex51
 2 years ago
Best ResponseYou've already chosen the best response.0Mertsj, how come the answer at the back fo my book is 7.8 degrees? o.o

milliex51
 2 years ago
Best ResponseYou've already chosen the best response.0ohh, never mind :$ THANK YOU!!
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