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Apr 2010
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Radio Station A is 100 miles due north of radio station B. Station A receives a distress message froma ship at a bearing of S 50° E, while station B receives the same message at a bearing of N 46° E.

How would a diagram for this look like?
 
Dec 2009
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Radio Station A is 100 miles due north of radio station B. Station A receives a distress message froma ship at a bearing of S 50° E, while station B receives the same message at a bearing of N 46° E.

How would a diagram for this look like?
S \(\displaystyle 50^o\) E means "point South, then turn to the East 50 degrees".
 

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Apr 2010
40
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Find the ratio of the sides of a triangle if:
(a) the angles are in the ratio 1:2:3
(b) the angles are in the ratio 4:5:6
 
Dec 2009
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Find the ratio of the sides of a triangle if:
(a) the angles are in the ratio 1:2:3
(b) the angles are in the ratio 4:5:6
Hi sinjid9,

if the ratio of angles is 1:2:3 there are 6 parts and the sum of the angles in a triangle is 180 degrees.

One angle is \(\displaystyle \frac{180^o}{6}=30^o\)

The other angles are twice and three times that, \(\displaystyle 60^o,\ 90^o\)

To calculate the ratios of the sides, you can re-arrange the Sine Rule (Law of Sines).

\(\displaystyle \frac{SinA}{a}=\frac{SinB}{b}=\frac{SinC}{c}\)

to get

\(\displaystyle \frac{SinA}{SinB}=\frac{a}{b}\)

then a:b=SinA:SinB

Also \(\displaystyle \frac{SinB}{b}=\frac{SinC}{c}\)

\(\displaystyle \frac{SinB}{SinC}=\frac{b}{c}\)

then a:b:c=SinA:SinB:SinC


When the ratio of the angles is 4:5:6, there are 15 parts

\(\displaystyle \frac{180^o}{15}=12^o\)

Then the angles are \(\displaystyle 4(12^o),\ 5(12^o),\ 6(12^o)\)

then solve as above
 
Apr 2010
40
0
A satellite travels in a circular orbit 1600km above Earth and takes 2 hours to complete one orbit. It will pass directly over a tracking station at 12:00 noon. If the tracking station can first pick up the satellite at a 30 degrees angle of elevation, at what time will the satellite appear? (The radius of the Earth is about 6400 km)

Does it look something like this?
http://img231.imageshack.us/i/mathpic2.png/
 

Soroban

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Hello, sinjid9!

You should start another thread for a new problem.


A satellite has a circular orbit 1600km above Earth and has a 2-hour orbit.
It will pass directly over a tracking station at 12:00 noon.
If the tracking station can first pick up the satellite at a 30 degrees angle of elevation,
at what time will the satellite appear? (The radius of the Earth is about 6400 km.)
Code:
    S
     o
      \ *
       \  *
   1600 \   *
         \ 30°* T
  H - - - \ - - o - - - - -
          *\    :     *
        *   \   :       *
       * 6400\  :6400    *
              \ : 
      *        \:         *
      * - - - - o - - - - *
                C

The center of the earth is \(\displaystyle C.\)
The tracking station is at \(\displaystyle T\!:\;TC = 6400\) km.
The satellite is at \(\displaystyle S\!:\;SC = 8000\) km.

The angle of elevation of the satellite is 30°.
. . \(\displaystyle \angle STH = 30^o,\;\angle STC = 120^o\)

In \(\displaystyle \Delta STC\), Law of Sines: .\(\displaystyle \frac{\sin S}{6400}\:=\:\frac{\sin120^o}{8000} \quad\Rightarrow\quad \sin S \:=\:0.692830323 \)

. . Hence: .\(\displaystyle \angle S \:\approx\:43.85^o\)

. . And: .\(\displaystyle \angle C \:=\:180^o - 120^o - 43.85^o \:=\:16.15^o\)


The satellite moves 360° in 2 hours.

To move 16.15°, it will take: .\(\displaystyle \frac{16.15^o}{360^o}\times 2 \;=\;\frac{323}{3600}\text{ hours} \;=\;\frac{323}{60}\text{ minutes}\)


The satellite will appear 5 minutes, 23 seconds before 12 noon.

 
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