IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => IGCSE/ GCSE => Math => Topic started by: spimuch on November 26, 2010, 09:09:13 pm
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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
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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
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Thanks dude =D
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another question :D..
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 .
(http://i154.photobucket.com/albums/s260/hilium/Untitled-5.png)
it is not a right angle triangle and i couldnt get and other triangle outside :\ ...
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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