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Trigonometry - Class 9 Problem

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The Whispered Secret

PostPosted: Wed Feb 11, 2009 12:40 pm


Hey theeeare =D

So, I have some Math homework. And
I have no idea how to do one of the questions.
So I was wondering if anyone would kindly help
me out? whee You'll be eternally loved.

So, the question is:

Quote:
A girl standing on a cliff top at A can see two
buoys X and Y, 200m apart, floating on the sea.
The angle of depression of X from A is 60 degrees
(see diagram)
If AXY are on the same vertical plane, calculate:
a) the distance AY
b) the distance AX
c) the vertical height of the cliff.

User Image


Thanks in advance. Please help. =3
xx
 
PostPosted: Sun Mar 15, 2009 3:02 pm


Ok, I'm labeling the point where the cliff meets the sea as C.

So angle CAX is 30 (90-60).

So the triangle between the cliff and the sea is a 30-60-90 right triangle.

Which means angle AXC is 60.

So angle AXY is 120 (180-60), angle XAY is 15 (60-45) and angle AYX is 45 (180-120-15).

Use the Law of Sines to get AX (sin(A)/a = sin(B)/b = sin(C)/c).

AX is the hypotenuse of the 30-60-90 triangle. So (AX/2)sqrt(3) is the height of the cliff.

Mooby the Golden Sock

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Mathematics

 
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