Bearing...1

[nycmathguy]

Section 4.8

I have lots of trouble understanding and setting up bearing problems. For some reason the wording is unclear to me.

I would like you to fully work out 33 and 34 (parts a & b) as a guide for me to solve similar problems.

 

[MathLover1]

33.
An airplane flying at 550mli per hour has bearing of 52 degrees. After flying for 1.5 hours, how far north and how far east will the plane have traveled from its point of departure?

t=1.5 h
s=550(mil/h)

distance= s*t
distance= 550(mil/h)*1.5 h=825mil

bearing of 52 degrees
so, the length to north, the length to east, and travel trajectory form right triangle where
the length to north =y
the length to east=x
travel trajectory=hypotenuse=825mil

then
a.
sin(52) = y/825 = 650mil N
b.
cos(52) = x/825 = 508mil E

 

[nycmathguy]

33.
An airplane flying at 550mli per hour has bearing of 52 degrees. After flying for 1.5 hours, how far north and how far east will the plane have traveled from its point of departure?

t=1.5 h
s=550(mil/h)

distance= s*t
distance= 550(mil/h)*1.5 h=825mil

bearing of 52 degrees
so, the length to north, the length to east, and travel trajectory form right triangle where
the length to north =y
the length to east=x
travel trajectory=hypotenuse=825mil

then
a.
sin(52) = y/825 = 650mil N
b.
cos(52) = x/825 = 508mil E

Let's put a hold on bearing problems FOR NOW..I want to watch a few math video clips on YouTube and get familiarized with the concept of beating. Be back tomorrow and Sunday with bearings.

 

[nycmathguy]

33.
An airplane flying at 550mli per hour has bearing of 52 degrees. After flying for 1.5 hours, how far north and how far east will the plane have traveled from its point of departure?

t=1.5 h
s=550(mil/h)

distance= s*t
distance= 550(mil/h)*1.5 h=825mil

bearing of 52 degrees
so, the length to north, the length to east, and travel trajectory form right triangle where
the length to north =y
the length to east=x
travel trajectory=hypotenuse=825mil

then
a.
sin(52) = y/825 = 650mil N
b.
cos(52) = x/825 = 508mil E

Nice work as always.

 

[nycmathguy]

33.
An airplane flying at 550mli per hour has bearing of 52 degrees. After flying for 1.5 hours, how far north and how far east will the plane have traveled from its point of departure?

t=1.5 h
s=550(mil/h)

distance= s*t
distance= 550(mil/h)*1.5 h=825mil

bearing of 52 degrees
so, the length to north, the length to east, and travel trajectory form right triangle where
the length to north =y
the length to east=x
travel trajectory=hypotenuse=825mil

then
a.
sin(52) = y/825 = 650mil N
b.
cos(52) = x/825 = 508mil E

I will tackle two more bearing two more bearing problems today. Before moving on to Chapter 5 (Analytic Trigonometry), there is another topic in Section 4.8, which I've never taken the time to learn. I am talking about Simple Harmonic Motion. I know trigonometry is involved. The topic is in most precalculus textbooks but for some reason, I never cared about it.