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January 30, 2015

January 30, 2015

Posted by **Jaskiran** on Sunday, January 29, 2012 at 10:15am.

a) The distance from point C to A

b) The bearing at which the pilot will have to fly to return to point A.

- Mathematics -
**Henry**, Monday, January 30, 2012 at 10:54pmd1 = 210km/h * 2h = 420 km @ 40 Deg.

d2 = 210km/h * 0.5h = 105 km @ 125 Deg.

X=hor.=420*cos40 + 105*cos125 = 262 m.

Y=ver.=420*sin40 + 105*sin125 = 356 m.

tanA = Y/X = 356 / 262 = 1.35878.

A = 53.6 Deg.

a. d = X/cosA = 262 / cos53.6 = 442 m =

Dist. from C to A.

b. Bearing = 53.6 Deg. North of East or

36.4 Deg. East of North.

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