1. ## Trig help

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?

2. Originally Posted by sinjid9
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 $50^o$ E means "point South, then turn to the East 50 degrees".

3. 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

4. Originally Posted by sinjid9
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 $\frac{180^o}{6}=30^o$

The other angles are twice and three times that, $60^o,\ 90^o$

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

$\frac{SinA}{a}=\frac{SinB}{b}=\frac{SinC}{c}$

to get

$\frac{SinA}{SinB}=\frac{a}{b}$

then a:b=SinA:SinB

Also $\frac{SinB}{b}=\frac{SinC}{c}$

$\frac{SinB}{SinC}=\frac{b}{c}$

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

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

$\frac{180^o}{15}=12^o$

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

then solve as above

5. 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/

6. 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 $C.$
The tracking station is at $T\!:\;TC = 6400$ km.
The satellite is at $S\!:\;SC = 8000$ km.

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

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

. . Hence: . $\angle S \:\approx\:43.85^o$

. . And: . $\angle C \:=\:180^o - 120^o - 43.85^o \:=\:16.15^o$

The satellite moves 360° in 2 hours.

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

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