Trig 11 Experts, please help!!?

Question

Can anyone please help me with this Trig question?

I'm cramming for a test tommorow morning and I'm lost...

ok, here goes:

In Extremely large forests, it is not cost effective to position forest rangers in towers or to use small aircraft to continually watch for fires. Since lightning is a frequent cause of fire, lightning detectors are now used instead.

A detector at point Q is situated 15 mi west of a central fire station at point R.
The bearing from Q to where lightning hits due south of R is South37.6(degrees)East.
How far is the hit from point R?

Please explain as much as possible*

Thanks a million!!!!!!!!

Answer 1

cos 37.6 degree = 15/distance
now find the value of cos
so distance to be found is 15/the value of cos 37.6 degree

Answer 2

I will assume that this means 127.6 degrees clockwise from North. I have a little trouble with South37.6East. Does this mean 37.6 degrees southward from east or 37.6 degrees eastward from south. In any case I have worked both out.

I will call the lightening point L.

Since Q is due west of R and L is due south of R the lines QR and RL intersect in a right angle. This means that QRL is a right triangle.

You know that QR = 15 miles and you know that the angle at Q is 37.6 degrees (if what I said above is correct). You want to find the distance the lightning is from R and this is RL. Just take the tangent of the angle at Q and that will be RL/QR or:

tan(Angle) = RL/QR = RL/15 = tan(37.6)
So RL = 15*tan(37.6) =11.55 miles

Note: If I have the angle wrong then it would be:
RL = 15*tan(52.4) = 19.48 miles

Answer 3

There are 3 points to be determined. Q - the detector position, R - the central fire station.
and let's say P is the location of the fire.
QPR is a triangle.
\We know that Q is due west of R, while P is due south of R. so QR is perpendicular to RP. QPR is a right angle triangle.
The bearing of QP (an imaginary line that connects the detector to the fire) is 37.6 degrees so that the the angle PQR=37.6 degrees.
We are looking for the distance PR.
We know that the distance QR is 15 miles.
The definition of the trigonometric function TAN in this right angle triangle is:
TAN(PQR)=PR/QR => PR=TAN(PQR)*QR
PR=TAN(37.6)*15=11.55 miles

Answer 4

The given archives defines a triangle ABC with C on the tip of a vertical line from A (2nd airport, C, is 230 km north of first, A) The beginning is factor A. The airplane leaves A for B at an perspective of eighty 5 deg. on the factor B,the place the airplane turns in direction of C the heading adjustments from eighty 5 to 283deg. This makes perspective ABC seventy seven deg (360 - 283). because of the fact the enclosed angles in a triangle sum to a hundred and eighty deg perspective BCA might desire to be 18 deg (a hundred and eighty-seventy seven-eighty 5). From the Sine rule BC/SinA = AC/SinB = AB/SinC, So (230/Sin77)=BC/Sin85, or (230/Sin77)*Sin85= BC = 235.15km. additionally (230/Sin77)=AB/Sin18 or (230/Sin77)*Sin18= AB = seventy two.ninety 4 km. So the completed is 230+235.15 +seventy two.ninety 4 = 538.09 Draw the triangle first then examine my figures. wish this helps in simple terms re-examine the question and can desire to be the respond is 308.09, this is the area the airplane flew from one airport to the subsequent. initially I secure traveling returned to its beginning (greater 230km)

Answer 5

Refer to this diagram: http://img519.imageshack.us/img519/4204/anszl7.png

By Alternate Angle Theorem,
Angle RLQ = 37.6°

tan RLQ = 15 / D
D = 15 / tan RLQ = 15 / tan 37.6° = 19.47789... = 19.5 mi approximately