Two Patrol boats M3 and M7 leave port at the same time. M3 heads due West and M7 on the bearing 277 degrees. After 30 minutes M7 has travel 18 meters and observe M3 in the direction due south.
a. How far is M3 from M7?
b. How far has M3 travelled?
a. To find the distance between M3 and M7, we can use trigonometry. Draw a diagram with M7 at the origin and M3 to the west. Then draw a line from M7 to M3, which represents the distance between them. We can use the angle between this line and the x-axis (which is 180 - 277 = 103 degrees) and the length of the line that M7 has traveled (18 meters) to find the length of the line connecting M3 and M7:
cos(103) = x/18
x = 18*cos(103) = 2.11 meters
So M3 is 2.11 meters away from M7 after 30 minutes.
b. M3 has traveled for 30 minutes due west. We can use the formula d = rt, where d is the distance traveled, r is the rate (speed), and t is the time. Since M3 is traveling due west, we know its direction is 270 degrees. We also know that the speed of both boats is the same, since they are leaving port at the same time. So:
d = r*t = speed * 0.5 hours = speed * 30 minutes
We just need to find the value of the speed. To do this, we can use the Pythagorean theorem and the distance we found in part a:
speed = sqrt(distance^2 + (18/2)^2) / 0.5
= sqrt(2.11^2 + 9^2) / 0.5
= 38.8 meters per hour
So M3 has traveled 38.8 * 0.5 = 19.4 meters in 30 minutes.
The bearing of a point A from P is 360 degrees and the bearing of another point B from P is 290 degrees. Both A and B due north of a point C which is 7600 metres due west of P. Find the distance between A and B.
First, let's draw a diagram:
```
A
|
|
|
C ----------------- P
|
|
|
B
```
We can see that AP is due north of C, so angle APC is a right angle. Similarly, BP is due north of C, so angle BPC is a right angle. Also, we know that CP is 7600 meters long.
Let's find the length of AP:
cos(APC) = adjacent/hypotenuse
cos(360) = AP / CP
1 = AP / CP
AP = CP = 7600 meters
Now let's find the length of BP:
cos(BPC) = adjacent/hypotenuse
cos(290) = BP / CP
BP = CP * cos(290) = 7600 * cos(290) = 678.4 meters
Finally, we can use the Pythagorean theorem to find the distance between A and B:
distance^2 = AP^2 + BP^2
distance = sqrt(AP^2 + BP^2) = sqrt(7600^2 + 678.4^2) = 7604.9 meters
Therefore, the distance between A and B is approximately 7604.9 meters.
Two walkers set off at the same time from a crossroad and walk along flat straight roads inclined to each other at 68 degrees. If they both walk at a speed of 6km/hr, find their distance apart 10 minutes later.
Let's start by drawing a diagram to visualize the problem.
```
A ------------- X ------------- B
/
/
/
Z
```
Let A and B be the starting points of each walker, X be the point where the two roads meet, and Z be the point where the two walkers are after 10 minutes. We want to find the distance between point A and point B, which is the distance that has opened up between the two walkers in 10 minutes.
We can use trigonometry to solve this problem. First, let's find the distance each walker has traveled in 10 minutes:
distance = rate * time = 6 km/hour * (10/60) hour = 1 km
Now, let's use the law of cosines to find the distance between points A and Z:
AZ^2 = XZ^2 + AX^2 - 2(XZ)(AX)(cos(68))
AZ^2 = (1 km)^2 + (1 km)^2 - 2(1 km)(1 km)(cos(68))
AZ^2 = 0.0785 km^2
AZ = sqrt(0.0785) km
AZ = 0.28 km (rounded to two decimal places)
Similarly, we can use the law of cosines to find the distance between points B and Z:
BZ^2 = XZ^2 + BX^2 - 2(XZ)(BX)(cos(68))
BZ^2 = (1 km)^2 + (1 km)^2 - 2(1 km)(1 km)(cos(68))
BZ^2 = 0.0785 km^2
BZ = sqrt(0.0785) km
BZ = 0.28 km (rounded to two decimal places)
Finally, we can find the distance between points A and B by adding AZ and BZ:
AB = AZ + BZ = 0.28 km + 0.28 km = 0.56 km
Therefore, the two walkers are approximately 0.56 km apart after 10 minutes.
Two patrol boats M3 and M7 leave port at the same port at the same time. M3 heads due West and M7 on the bearing 277 degrees. After 30 minutes M7 travel 18 meters and observes M3 is the direction due south.
a. How far is M3 from M7
b. How far has M3 travelled ?
Let's start by drawing a diagram to visualize the problem:
```
M3 ---------*-------
| /
| /
| /
| /
|/
M7 ---------*--------
```
We can see that M7's direction is 277 degrees, which means that the angle between M7's direction and due north is 360 - 277 = 83 degrees. We also know that M7 has traveled for 30 minutes at a speed of 18 meters. We can use this information to find the position of M7 after 30 minutes and the distance between M7 and M3.
To find the position of M7 after 30 minutes, we can use the formula d = rt, where d is the distance traveled, r is the rate (speed), and t is the time:
distance = rate * time = 18 meters
Now, we can use trigonometry to find the distance between M7 and M3:
tan(83) = opposite/adjacent
opposite = adjacent * tan(83)
We know that M7 has traveled a distance of 18 meters, so we can use Pythagoras' theorem to find the distance between M7 and M3:
distance = sqrt((18*cos(83))^2 + (18*sin(83) + x)^2)
We don't know the value of x yet, which represents the distance that M3 has traveled. To find it, we can use the fact that M3 is due south of M7 after 30 minutes, which means that the distance between M3 and the point directly south of M7 must also be 18 meters (since M7 has traveled 18 meters). Let's call this point Y:
M3 ---------*-------
| /
| /
| /
| / Y
|/
M7 ---------*--------
We can use trigonometry to find the distance between M3 and Y:
cos(83) = opposite/adjacent
opposite = adjacent * cos(83)
Now, we can use Pythagoras' theorem to find x:
x^2 = (18*sin(83) + opposite)^2
x^2 = (18*sin(83) + adjacent*cos(83))^2
x = 5.2 meters (rounded to one decimal place)
Therefore, the two answers are:
a. The distance between M3 and M7 is approximately 18.3 meters (rounded to one decimal place).
b. The distance traveled by M3 is approximately 5.2 meters (rounded to one decimal place).
The bearing of a point A from P is 330 degrees and the bearing of another point B from P is 290 degrees. Both A and B due north of a point C which is 7600 meters due West of P. Find the distance between A and B.
Let's start by drawing a diagram to visualize the problem:
```
A
|
|
|
C -------------------- P
/
/
/
|
|
B
```
We can see that A and B are both due north of point C, and that PC is 7600 meters due west of P. To find the distance between A and B, we need to find the length of AC and the length of BC.
To find the length of AC, we can use trigonometry. The bearing of point A from P is 330 degrees, which means that the angle between PA and due north is 360 - 330 = 30 degrees. Therefore, we have:
tan(30) = opposite/adjacent
opposite = adjacent * tan(30)
We know that PC is 7600 meters, so AC must be (7600 * tan(30)) meters:
AC = 7600 * tan(30)
AC = 4387.7 meters (rounded to one decimal place)
Similarly, to find the length of BC, we can use the bearing of point B from P, which is 290 degrees. The angle between PB and due north is 360 - 290 = 70 degrees, so we have:
tan(70) = opposite / adjacent
opposite = adjacent * tan(70)
We know that PC is 7600 meters, so BC must be (7600 * tan(70)) meters:
BC = 7600 * tan(70)
BC = 2273.4 meters (rounded to one decimal place)
Now we can use the Pythagorean Theorem to find the distance between A and B:
distance^2 = AC^2 + BC^2
distance = sqrt(AC^2 + BC^2)
distance = sqrt((4387.7)^2 + (2273.4)^2)
distance = 5007.5 meters (rounded to one decimal place)
Therefore, the distance between points A and B is approximately 5007.5 meters.