Math Help - geometry test

1. geometry test

hello i have a major test in a couple of days and i want to go over some geometry.
q1) how far apart are Adelaide 35S, 139E and Tokyo 36N, 139E?
add the latitute so 35S plus 36 N=71
71 divded 360 times 2 times pie times 6400=7930.78km
nautical miles=7931km divided by 1.852=
4282 nm.

two places on the equator have longitude 150e, 20 w.Find the shortest distance between these two places.correct to the nearest km. 150 plus 20 =170.
170 divded by 360 times 2 times pie times 6400=18989km to nearest km.

fred lives in adelaide (35,s 139E) he wants to watch footbal in spain (37n 6W) starting 2pm sunday. what time is it in Adelaide when the football starts?

A yacht leaves naura 1S, 167E and sails to fiji 16S, 167E.The yaht leaves nauru at 5pm thurday 3rd jan.

find the distance in km between these to places.
I get confused with nautical miles and km.

the yaht sales an average of 8 knots. how many hours from naura to fiji?:

give the date and time of arrival of the yaht in fiji?

2. Originally Posted by mat
...
(1) fred lives in adelaide (35,s 139E) he wants to watch footbal in spain (37n 6W) starting 2pm sunday. what time is it in Adelaide when the football starts?

(2) A yacht leaves naura 1S, 167E and sails to fiji 16S, 167E.The yaht leaves nauru at 5pm thurday 3rd jan.

find the distance in km between these to places.
I get confused with nautical miles and km.

the yaht sales an average of 8 knots. how many hours from naura to fiji?:

give the date and time of arrival of the yaht in fiji?
to (1):

Keep in mind that Adelaide is located East of Spain, that means the sun rises first in Adelaide and later in Spain.
The azimut between Spain and Adelaide is 145°. Since an hour correspond to 15° the sun rises nearly 10 hours earlier in Adelaide than in Spain. So it is nearly midnight in Adelaide when the play starts in Spain. (Maybe you should get some informations about the correct time zones)

to (2):

As I've posted you can use the property: 1° on a great circle correspond with 60 nm (=60'). Since the yacht sails on a great circle it has covered the distance of 15° = 900' = 900 nm.

Now convert the nautical miles into kilometres if necessary: $900\ nm \cdot 1.852\ \frac{km}{nm} \approx 1666.8\ km$

$1\ kt = 1\ \frac{nm}{h}$

From $speed=\dfrac{distance}{time}$ follows $time=\dfrac{distance}{speed}$

Therefore the yacht needs: $t=\dfrac{900\ nm}{8\ h}= 112.5\ h$

Convert into days and hours and calculate the date and time of the arrival.

3. just couple more questions if i get 154 degrees divided by 15 =8.26 this is 8 hours and 26 minutes correct? what about 112.5 hours is 112 hours and 50 minutes and 9.86 is that 10 hours and 26 minutes? thanks.

4. Originally Posted by mat
just couple more questions if i get 154 degrees divided by 15 =8.26 this is 8 hours and 26 minutes correct? what about 112.5 hours is 112 hours and 50 minutes and 9.86 is that 10 hours and 26 minutes? thanks.
1. If you calculate

$\dfrac{154}{15}\approx 10.2667$

which is the same as 10 hours and $\dfrac4{15}$ hours. Since \dfrac1{15} hour correspond to 4 min the the period of time of

$\dfrac{154}{15}=\left(10 + \dfrac4{15}\right)\ h=10\ h + 16\ min$

2. Keep in mind that 0.1 h = 6 min. Therefore 112.5 h = 112 h + 30 min. By the way: You never would say that half an hour is the same as 50 min (only in school when you are too late )