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.
This paragraph is nonsense:
"We can also use the fact that both A and B are due north of C. This means that the distance between A and C, as well as the distance between B and C, is equal to the distance between P and C, which is 7600 meters."
We cannot say that at all.
What we can say is that
BC/7600 = tan20°
AC/7600 = tan60°
That gives us
AB = AC-BC = 7600(tan60°-tan20°) = 10397 meters
You are correct, I apologize for the error in my explanation. Your solution is correct. Thank you for pointing out my mistake.
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.
We can start by drawing a diagram of the situation. Let's call the two walkers A and B, and let's say that they start at point O:
A
|
| 68°
|
|_____________
O B
We know that the angle between the two roads (OA and OB) is 68 degrees. We also know that both walkers walk at a speed of 6 km/hr.
After 10 minutes, A would have covered a distance of (6/60)*10 = 1 km. Similarly, B would have covered a distance of 1 km. Let's call the position of A after 10 minutes A', and the position of B after 10 minutes B':
A'
|
| 68°
|
|_____________
O B'
We want to find the distance between A' and B'. To do this, we can use the Law of Cosines. Let's call the distance between A and O "a," the distance between B and O "b," and the distance between A' and B' "c."
From the diagram, we see that angle A'OB' is 112 degrees (180 - 68 - 68). Using the Law of Cosines, we have:
c² = a² + b² - 2ab*cos(112)
c² = a² + b² + 2ab*cos(68)
We know that a = b = 1 km, so:
c² = 2 - 2*cos(68)
c = sqrt(2 - 2*cos(68))
c ≈ 1.94 km
Therefore, the distance between A' and B' after 10 minutes is approximately 1.94 km.
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.
We can start by drawing a diagram of the situation. Let's call the two walkers A and B, and let's say that they start at point O:
A
|
| 68°
|
|_____________
O B
We know that the angle between the two roads (OA and OB) is 68 degrees. We also know that both walkers walk at a speed of 6 km/hr.
After 10 minutes, A would have covered a distance of (6/60)*10 = 1 km. Similarly, B would have covered a distance of 1 km. Let's call the position of A after 10 minutes A', and the position of B after 10 minutes B':
A'
|
| 68°
|
|_____________
O B'
We can now use the Law of Cosines to find the distance between A' and B'. Let's call this distance "d." We know that:
- Angle A'OB' is 112 degrees (180 - 68 - 68)
- Side OA' (and OB') has length 1 km
- Both sides OA' and OB' make an angle of 68 degrees with side AB (hypothenuse of triangle AOB)
Using the Law of Cosines, we get:
d^2 = 1^2 + 1^2 - 2(1)(1)cos(112)
d^2 = 2 - 2cos(112)
We can use the fact that cos(112) = -cos(68) to simplify this expression:
d^2 = 2 + 2cos(68)
d ≈ 2.45 km
Therefore, the two walkers will be approximately 2.45 km apart after 10 minutes.
A map has a scale of 1:50000. Two towers are 10.4 km apart. How many centimeters on the map is this
We know that 1 cm on the map represents 50,000 cm (or 500 m) in real life. We want to find out how many centimeters on the map represent a distance of 10.4 km.
We can set up a proportion:
1 cm / 50,000 cm = x cm / 10,400,000 cm
where x is the number of centimeters on the map that represents a distance of 10.4 km.
We can cross-multiply:
50,000 * x = 10,400,000
x = 208 cm
Therefore, 10.4 km on the ground is represented by 208 cm on the map.