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?
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
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/
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.