Sunday

March 1, 2015

March 1, 2015

Posted by **Dakota S** on Sunday, September 12, 2010 at 9:08pm.

- Physics -
**drwls**, Monday, September 13, 2010 at 1:38am(a) Add the vectors of his 55 mile and 25 miles flights. The reverse of that vector is the direction and distance he must fly the next day.

(b) Add the magnitudes of the three flight vectors in the previous question.

- Physics -
**Henry**, Monday, September 13, 2010 at 2:41am15 deg. E of S = 285 deg CCW from 0 deg

X = HOR = 25cos(285) - 55 = -48.5,

Y = VER = 25sin(285) = -24.15,

tanA = Y/X = -24.15/-48.5 = 0.4979,

A = 26.47 deg. = 206.47 deg.(3rd quad.)

d = Y/sinA = -24.15/sin(206.5)=54.18mi

d = 54.18 mi @ 206.47 deg.

a. 54.18 mi N @ 63.5 deg W of South

Location: 3rd QUADRANT.

B. D = 2 * 54.18 = 108.36 MILES

**Answer this Question**

**Related Questions**

Trig - A plane is 48 miles west and 49 miles north of an airport. The pilot ...

math - A plane is 11.9 miles off the ground as it makes its descent to land at ...

Math - A plane flies due east for 120 km from airport a to airport b it then ...

trig - A plane is flying at an elevation of 30000 feet. It is within sight of ...

Pre Cal - An airplane leaves an airport and flies due west 150 miles and then ...

Math - An airplane leaves the airport and flies due west 170 miles and then 240 ...

trig - an airplane leaves the airport and flies due west 150 miles and then 240 ...

Math/Trig - An airplane leaves an airport and flies due west 150 miles and then ...

Trigonometry - A plane is 160 miles north and 85 miles east of an airport. If ...

Trigonometry - Round answer to two significant digits. An airplane flew 300 ...