Math question ...ans plz exam tommorow :)

[spimuch]

i had problem solving the following questions and i have exams tommorow ...

An aircraft flies a certain distance on a bearing of 135 degrees and then twice the distance on a bearing of 225 degrees .its distance from the starting point is then 350 km.find the lenth of the first part of the journey .

my answer using the formula is 167 ...
and the answer in the book is 157 .

Thanks =D

[Freaked12]

If you draw the graph

you will see we have 3 lengths
x, 2x (twice the distance) and 350

and since this is a right-angled triangle ,we can use the Pythagoras theorem to determine the value of x.

x^2+(2x)^2=350^2

5x^2=122500
x^2=24500
taking the under root we get x=156.5 or 157

[spimuch]

Thanks dude =D

[spimuch]

another question ..

An aircraft flies 500 KM on a bearing of 100 degrees and then 600km on bearing of 160 degrees.find the distance and bearing of the finishing point from the starting point .




it is not a right angle triangle and i couldnt get and other triangle outside :\ ...

[Freaked12]

You seem to be mistaken

You can use the cosine rule (to get the distance) and sine rule(to get the angle) in any triangle

a^2=b^2+c^2-2bc Cos(A)
a^2=(500)^2+(600)^2-2(500)(600) cos(120)
a^2=910000
a=953.9 Km

Sin A     Sin B
------= ------
   A          B

Sin (120)    Sin(x)
---------= -------
    953.9       500

you will get x=27
you add it with 20(which you got by comparing parallel lines) and you get 47.
360-47=313 degree

Hope that helps