# Trig help

#### 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?

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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#### 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

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

#### sinjid9

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

MHF Hall of Honor
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.

• sinjid9