A tropical location B is 20 km N15° E of A. Another location C is N45°W of A and S75° W of B. How far (in km) is C from A? (N15° E denotes a direction 15° east of north.)
20
Tropical Location Distance Calculation
The problem describes the relative positions of three tropical locations, A, B, and C, using distances and bearings. We are given the distance from A to B and the bearings of B from A, C from A, and C from B. We need to find the distance from C to A.
Understanding the Given Information
- Location B is 20 km from A in the direction N15° E.
- Location C is in the direction N45°W of A.
- Location C is in the direction S75°W of B.
We can represent the locations A, B, and C as vertices of a triangle. The distances are the lengths of the sides, and the bearings help us determine the angles of the triangle.
Calculating Angles of Triangle ABC
Let's determine the angles within the triangle ABC.
- Angle at A (∠CAB):
- The direction of AB from A is N15°E. This means the angle between the North direction at A and the line segment AB is $15^\circ$.
- The direction of AC from A is N45°W. This means the angle between the North direction at A and the line segment AC is $45^\circ$.
- Since N15°E is East of North and N45°W is West of North, the angle between AB and AC is the sum of these angles: ∠CAB = $15^\circ + 45^\circ = 60^\circ$.
- Angle at B (∠ABC):
- The direction of AB from A is N15°E. The reciprocal bearing (direction of BA from B) is S15°W. This means the angle between the South direction at B and the line segment BA is $15^\circ$.
- The direction of BC from B is S75°W. This means the angle between the South direction at B and the line segment BC is $75^\circ$.
- Since both directions (BA and BC) are West of South, the angle between BA and BC is the difference between their angles from South: ∠ABC = $75^\circ - 15^\circ = 60^\circ$.
- Angle at C (∠BCA):
- The sum of angles in a triangle is $180^\circ$.
- ∠BCA = $180^\circ - \angle CAB - \angle ABC$.
- ∠BCA = $180^\circ - 60^\circ - 60^\circ = 60^\circ$.
Identifying the Triangle Type
We found that all three angles of triangle ABC are $60^\circ$ (∠CAB = $60^\circ$, ∠ABC = $60^\circ$, ∠BCA = $60^\circ$).
A triangle with all three angles equal to $60^\circ$ is an equilateral triangle.
Calculating the Distance from C to A
In an equilateral triangle, all sides are equal in length.
We are given that the distance from A to B (side AB) is 20 km.
Since triangle ABC is equilateral, the length of side AC and side BC must also be equal to the length of side AB.
Therefore, the distance from C to A (side AC) is 20 km.
Final Distance
The distance from C to A is 20 km.
Important Questions from Miscellaneous
A stone is thrown horizontally from the top of a 20 m high building with a speed of 12 m/s. It hits the ground at a distance R from the building. Taking g = 10 m/s2 and neglecting air resistance will give :
A sphere of volume V is made of a material with lower density than water. While on Earth, it floats on water with its volume f1V (f1 < 1) submerged. On the other hand, on a spaceship accelerating with acceleration a < g (g is the acceleration due to gravity on Earth) in outer space, its submerged volume in water is f2V. Then:
A railway wagon (open at the top) of mass M1 is moving with speed v1 along a straight track. As a result of rain, after some time it gets partially filled with water so that the mass of the wagon becomes M2 and speed becomes v2. Taking the rain to be falling vertically and the water stationery inside the wagon, the relation between the two speeds v1 and v2 is :
Consider the following statements:
1. Distance between the longitudes becomes zero on North Pole and South Pole.
2. Distance between the longitudes is maximum on the Equator.
3. Number of longitudes is more than number of latitudes.
Which of the statements given above is/are correct?
One block of 2⋅0 kg mass is placed on top of another block of 3⋅0 kg mass. The coefficient of static friction between the two blocks is 0⋅2. The bottom block is pulled with a horizontal force F such that both the blocks move together without slipping. Taking acceleration due to gravity as 10 m/s2, the maximum value of the frictional force is :
