milliex51
How do I find the measure of the indicated angle, to the nearest tenth of a degree using sine law and cosine law?
Delete
Share
This Question is Closed
milliex51
Best Response
You've already chosen the best response.
0
|dw:1337042419972:dw|
jiberjiber
Best Response
You've already chosen the best response.
0
Uh, what?
jiberjiber
Best Response
You've already chosen the best response.
0
|dw:1337043771132:dw|
jiberjiber
Best Response
You've already chosen the best response.
0
four degrees?
milliex51
Best Response
You've already chosen the best response.
0
no, it's 104 degrees and the side length of BC is 45 m
Mertsj
Best Response
You've already chosen the best response.
1
|dw:1337046316750:dw|
Mertsj
Best Response
You've already chosen the best response.
1
Let's say that side AB is c and use the law of cosines.
milliex51
Best Response
You'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
Mertsj
Best Response
You've already chosen the best response.
1
\[c^2=20^2+45^2-2(20)(45)\cos 104\]
milliex51
Best Response
You'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
milliex51
Best Response
You've already chosen the best response.
0
and now I am stuck..
Mertsj
Best Response
You've already chosen the best response.
1
|dw:1337046557115:dw|
Mertsj
Best Response
You've already chosen the best response.
1
Now find the meassure of angle ACB using the law of sines.
Mertsj
Best Response
You've already chosen the best response.
1
|dw:1337046713091:dw|
milliex51
Best Response
You'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..
Mertsj
Best Response
You've already chosen the best response.
1
\[\frac{\sin 104}{53.48}=\frac{\sin ACB}{20}\]
Mertsj
Best Response
You've already chosen the best response.
1
Angle ACB = 21.3 degrees
milliex51
Best Response
You've already chosen the best response.
0
how did you get 21.3?
milliex51
Best Response
You've already chosen the best response.
0
and the angle of C has to be 7.3 degrees..
Mertsj
Best Response
You've already chosen the best response.
1
|dw:1337046878417:dw|
Mertsj
Best Response
You've already chosen the best response.
1
20(sin 104)/52.48
Then hit the inverse sin button.
milliex51
Best Response
You've already chosen the best response.
0
how did you get 54.7 and 21.3? :O i'm so sorry
Mertsj
Best Response
You'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
milliex51
Best Response
You've already chosen the best response.
0
ohh yup. But how am I going to find angle C? D:
Mertsj
Best Response
You've already chosen the best response.
1
But you want the angle theta so try this:
\[(CD)^2=30^2+45^2-2(30)(45)\cos 104\]
Mertsj
Best Response
You've already chosen the best response.
1
|dw:1337047600966:dw|
milliex51
Best Response
You've already chosen the best response.
0
i'm sorry but it just says, 'math processing error..'
Mertsj
Best Response
You've already chosen the best response.
1
|dw:1337047693118:dw|
Mertsj
Best Response
You'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.
milliex51
Best Response
You've already chosen the best response.
0
okay. I'm still kind of confused but thank you so much!
Mertsj
Best Response
You've already chosen the best response.
1
\[\frac{\sin 104}{59.8}=\frac{\sin DCB}{30}\]
Mertsj
Best Response
You've already chosen the best response.
1
What are you confused about?
milliex51
Best Response
You've already chosen the best response.
0
Mertsj, how come the answer at the back fo my book is 7.8 degrees? o.o
milliex51
Best Response
You've already chosen the best response.
0
ohh, never mind :$ THANK YOU!!