An airplane flies 165 miles from point A in the direction 130 degrees and then travels in the direction 245 degrees for 80 miles. Approximately how far is the airplane from A?

the first leg moves the plane in th x- and y-directions by

(165*cos40°,-165*sin40°)

Now figure the amounts of the second leg, add them up, and then use the distance formula.

The subject says algebra, but since you apparently have had some trig, just use the law of cosines:

x^2 = 165^2 + 245^2 - 2(165)(245)cos115°

I really don't know where Steve got his numbers from.

Making a sketch I have a triangle with sides 165 and 80 and an angle of 65° between them, so by the cosine law
x^2 = 80^2 + 165^2 - 2(80)(165)cos 65°
= 22467.87789
x = appr 149.9 miles

Using vectors we have
(165cos 30°, 165sin130°) + (80cos245°,80sin245°)
= ....
= (-139.869.. , 53.892..)
and the magnitude of that vector is 149.9 miles

check: the direction :
tanØ = 53.892/-39.869
= -.385...
Ø = 158.9 or appr 159° which was consisten with my diagram.

To find the approximate distance of the airplane from point A, we can use the concept of vector addition.

1. First, let's break down the two vectors into their horizontal (x) and vertical (y) components.

For the vector traveling in the direction of 130 degrees:
- The horizontal component (x) is calculated as 165 * cos(130 degrees).
- The vertical component (y) is calculated as 165 * sin(130 degrees).

For the vector traveling in the direction of 245 degrees:
- The horizontal component (x) is calculated as 80 * cos(245 degrees).
- The vertical component (y) is calculated as 80 * sin(245 degrees).

2. Next, we add the respective horizontal and vertical components together for each vector to get the resultant x and y components.

3. Finally, we can use the Pythagorean theorem to find the distance (d) of the airplane from point A, using the resultant x and y components:
d = sqrt(x^2 + y^2)

Performing the calculations, we can find the approximate distance.