trig problem: find approximate distance between kites

[kpx001]

how would i do this word problem??

a boy is flying two kits at the same time. he has 380 feet of line out to one kite and 420 feet to the other. he estimates the angle between the two lines to be 30 degrees. approximate the distance between the two kites to the nearest hundred.

i labed a picture and labeled 30 as A and 380 as side B and c as 420. from there i dunno if i have to do a law or something.

 

[skeeter]

you have a triangle with a known side-angle-side ...
use the law of cosines, \(\displaystyle a^2 = b^2 + c^2 - 2bc\cos(A)\)

 

[kpx001]

a^2 = b^2 + c^2 - 2bccos(A)

a^2 = 380^2 + 420^2 - 2(380)(420)cos30

1600cos30 = a^2

1600(2*sqroot(3)/3) = a^2

a = 42.98 ?? did i do anything wrong? or is this correct?

 

[skeeter]

a^2 = b^2 + c^2 - 2bccos(A)

a^2 = 380^2 + 420^2 - 2(380)(420)cos30

1600cos30 = a^2 incorrect from here on ...

1600(2*sqroot(3)/3) = a^2

a = 42.98 ?? did i do anything wrong? or is this correct?

 

[kpx001]

is it the cos value? when i plugged in calculator i got 37.22 as the answer

 

[Deleted member 4993]

a^2 = b^2 + c^2 - 2bccos(A)

a^2 = 380^2 + 420^2 - 2(380)(420)cos30

1600cos30 = a^2.............How did you get to this from above?

1600(2*sqroot(3)/3) = a^2

a = 42.98 ?? did i do anything wrong? or is this correct?

 

[skeeter]

do this on your calculator ...

\(\displaystyle \L a = \sqrt{380^2 + 420^2 - 2*380*420\cos(30)}\)

 

[kpx001]

210.63?

 

[skeeter]

210.63 ft

 

[stapel]

approximate the distance between the two kites to the nearest hundred.

Really? To the nearest "hundred"? Not the nearest "hundredth"? :shock:

Eliz.