Question 1194743: A certain kilometer post on a road running N 67°E is 2 kms. N 30°W of a police observation tower. Ten minutes after a car passes this kilometer post, it is seen to bear N 33°E of the tower. If the speed limit is 50 kph, was the car over speeding or not? How fast was he driving?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
The notation N 30° W means we first look directly north. Then turn 30 degrees to the west. This is shown in the diagram below as angle ACD (in blue). The reference point is point C.
The other bearing angles are handled in a similar fashion.
Diagram:
Points:
A = Kilometer post (where the car is first spotted)
B = Car's position after 10 minutes of passing point A
C = Police tower
D = Point directly north of point C
E = Point directly north of point A
F = Point directly south of point A
Segments:
AB = x
AC = 2
Angles:
Angle EAB = 67° (green)
Angle ACD = 30° (blue)
Angle BCD = 33° (red)
The goal is to find the length of segment AB, to find out how much distance the car traveled in those 10 minutes.
If we can find that distance, then we can determine the speed due to the formula:
speed = distance/time
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Angles CAF and ACD are alternate interior angles.
Because segment FA is parallel to segment CD, we know the alternate interior angles are congruent.
This means angle CAF is 30 degrees.
Angles EAB, BAC, and CAF form a straight angle of 180 degrees. Use this fact to determine angle BAC.
(angle EAB) + (angle BAC) + (angle CAF) = 180
(67) + (angle BAC) + (30) = 180
angle BAC + 97 = 180
angle BAC = 180 - 97
angle BAC = 83 degrees
If we focus solely on triangle ABC, then we can refer to "angle BAC" as "angle A".
Keeping our focus on just this triangle, the blue and red angles (ACD and BCD respectively) add to 30+33 = 63 degrees.
This is the measure of angle ACB, or in short, angle C.
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Let's summarize what we have so far that is relevant:
Focus entirely on triangle ABC. Erase/ignore points D,E,F since they aren't needed anymore.- Angle A = 83 degrees
- Angle C = 63 degrees
- Side AC = 2 km
Let's find angle B
A+B+C = 180
83+B+63 = 180
B+146 = 180
B = 180-146
B = 34 degrees
Notice that side AC is opposite angle B
I'll use lowercase b to indicate this side.
side b = AC = 2 km.
Now we'll use the law of sines to find side c (aka segment AB)
sin(B)/b = sin(C)/c
sin(34)/2 = sin(63)/x
x*sin(34) = 2*sin(63)
x = 2*sin(63)/sin(34)
x = 3.1867590545522 approximately
The driver covered roughly 3.1867590545522 km in 10 minutes. This timespan is equivalent to 10/60 = 1/6 of an hour.
We can now compute the average speed.
speed = distance/time
speed = (3.1867590545522 km)/( 1/6 of an hour )
speed = 19.1205543273132
The driver is traveling at a speed of about 19.12 kph, which is well under the 50 kph speed limit.
It might be tempting to say that this low speed isn't realistic, but the road/weather conditions may be hazardous that it's better to drive at a safer slower speed.
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