F3 三角學的應用 (好多日幫冇人覆)

12.
(a) Draw a line AC perpendicular to BC. The plane flies along BC. The shortest distance the plane to A is AC.
Angle ABC = 75 - 50 = 25
Sin 25 = AC/AB = AC/100 km
AC = 100 km sin 25 = 100(0.4226) = 42.26 km
Since the radar can detect plane flying within 50 km from A, therefore the plane can be detected.

(b) Draw a line ad such that AD = 50 km
Consider triangle ABD,using sine law
Sin 25/50 = sin ADB/ 100
Sin ADB = (100) Sin 25/50 = (100)0.4226/50 = 0.8452
ADC = 57.7 or 122.3 
ADC = 122.3 
Angle BAD = 180 - 122.3 - 25 = 32.7
Using sine law,
50/sin 25 = BD/ sin 32.7
BD = 50 sin 32.7/ sin 25 = 50(0.5402)/0.4226 = 63.92 km
The plane must fly a distance of 63.92 km before being detected
Time = distance/speed =63.92 km/150 km/h = 0.4261 h = 25 min. 34 s
The plane must fly 25 min 34 s before being detected.

12.
slope = 5/12, hypotenuse = 13
The horizontal distance (1.3 km/13)(12) = 1.2 km
1.2 cm represents 1.2 km
1 cm on the map represents 1 km
1 km = 1000 (100) = 100000 cm
1: n
n = 100000

33.
(a) Can’t draw on yahoo editor.
(b) Let the original place be A. B is 10 km north of A. C is the destination.
Consider triangle ABC. Angle ABC = 120 (90+30)
AB = 10 km, BC = 8 km
Using Cosine law
AC^2 = AB^2 + BC^2 – 2(AB)(BC) cos 120
AC^2 = 10^2 + 8^2 – 2(10)(8) cos 120
AC^2 = 164 – 160 (-0.5)
AC^2 = 164 + 80 = 244
AC = 244^0.5
AC = 15.62 km
(c)
Using sine law
Sin B/AC = sin A/BC
Sin 120/15.63 = sin A/8
Sin A = 8 sin120/15.63 = 0.4438
A = 26.35
N 26.35°E
(d) Either return in original path, from C to B, then from B to A
or go back directly straight from C to A

I haven’t got time to check my answers ~ mainly to show the steps.