Question

Two weather tracking stations are on the equator 146 miles apart. A weather balloon is located on a bearing of N 35°E from the western station and on a bearing of N 23°E from the eastern station. How far is the balloon from the western station?

Answer 1

you don't have any choices right?

Answer 2

no

Answer 3

all i have is a diagram attached above

Answer 4

ok

Answer 5

could you help me figure it out

Answer 6

\(\bf tan(55^o)=\cfrac{{\color{brown}{ y}}}{146+{\color{blue}{ x}}}\qquad tan(67^o)=\cfrac{{\color{brown}{ y}}}{{\color{blue}{ x}}}
\\ \quad \\\\ \quad \\ \quad \\
tan(55^o)=\cfrac{{\color{brown}{ y}}}{146+{\color{blue}{ x}}}\implies
(146+{\color{blue}{ x}})[tan(55^o)] = {\color{brown}{ y}}
\\ \quad \\

tan(67^o)=\cfrac{{\color{brown}{ y}}}{{\color{blue}{ x}}}\implies tan(67^o)\cdot {\color{blue}{ x}}={\color{brown}{ y\qquad thus }}
\\ \quad \\
{\color{brown}{ y}}={\color{brown}{ y}}\implies (146+{\color{blue}{ x}})[tan(55^o)] =tan(67^o)\cdot {\color{blue}{ x}}
\\ \quad \\
146+x=\cfrac{tan(67^o)}{tan(55^o)}\cdot x\)

Answer 7

im confused i thought i would use laws of sines for this problem where does tan come from

Answer 8

hmmm didn't know you were supposed to use the law of sines...

Answer 9

yea

Answer 10

ohhh shoot then that's simpler =)

Answer 11

i get how to use the laws of sines on this i just dont get how you got 67 and 55

Answer 12

please explain with laws of sines=)