1. The orbit of the earth is an ellipse with the sun at a focus, the semi-major axes 93 million miles and the eccentricity is 1/60. Find the greatest and least distance of the earth from the sun.
2. A battle shipping is sailing at a straight line towards east at 40km/hour from start and another smaller ship sailed north at 60km/hour. If both ship started at the same time, 2 hours later, the battleship is now at point C, the battleship wants to launch a missile towards B. If the smaller ship headed north (now at point A 2 hours later) found that point But is 10 degrees north of east and it is 150 km away. Solve the angle that the battleship should direct its missile with respect to point A in order to hit point B.
distance extremes are at the ends of the semi-axes. You know that a=93, and c/a = 1/60. And, b^c+c^2=a^2.
Hard to figure just what you are after with the ships, but apparently you want the angle ACB.
Angle CAB is arctan(120/80)+10°
To find the distance CB, use the law of cosines.
CB^2 = 150^2 + (40√13)^2 - 2(150)(40√13)cos(CAB)
Then using the law of sines,
sin(CAB)/150 = sin(CAB)/BC
1. To find the greatest and least distance of the Earth from the Sun, we can use the formula for distance in an ellipse:
For an ellipse with semi-major axis a and eccentricity e:
Greatest distance = a(1 + e)
Least distance = a(1 - e)
Given that the semi-major axis of the Earth's orbit is 93 million miles (a = 93 million miles) and the eccentricity is 1/60 (e = 1/60), we can substitute these values into the formulas:
Greatest distance = 93 million miles * (1 + 1/60)
Least distance = 93 million miles * (1 - 1/60)
Now let's calculate the distances:
Greatest distance = 93 million miles * (1 + 1/60)
= 93 million miles * (61/60)
≈ 94.45 million miles
Least distance = 93 million miles * (1 - 1/60)
= 93 million miles * (59/60)
≈ 91.85 million miles
Therefore, the greatest distance of the Earth from the Sun is approximately 94.45 million miles, and the least distance is approximately 91.85 million miles.
2. To solve for the angle that the battleship should direct its missile with respect to point A in order to hit point B, we can use trigonometry.
Let's denote the angle that the battleship should direct its missile with respect to point A as θ.
Using the given information, we can form a right triangle with sides BC, AC, and AB.
Since BC is 150 km and AB is 2 hours of travel time for the battleship at 40 km/hour (AB = 2 * 40 = 80 km), we have:
cos(θ) = BC / AB
cos(θ) = 150 km / 80 km
cos(θ) = 1.875
Now, to find θ, we can take the inverse cosine (arccos) of 1.875:
θ = arccos(1.875)
θ ≈ 47.658 degrees
Therefore, the battleship should direct its missile at an angle of approximately 47.658 degrees with respect to point A in order to hit point B.
1. To find the greatest and least distance of the earth from the sun, we can use the formula for the distance of a point on an ellipse from its center.
The distance from the sun to any point on the ellipse can be given by:
r = a(1 - e^2) / (1 + e*cos(theta))
Where:
- r is the distance from the sun to the point on the ellipse
- a is the semi-major axis length
- e is the eccentricity of the ellipse
- theta is the angle between the line connecting the sun and the point on the ellipse, and the major axis of the ellipse (where the reference angle is 0)
In this case, we are given:
- a = 93 million miles
- e = 1/60
To find the greatest and least distance, we need to find the values of r that correspond to the minimum and maximum values of the expression above.
The greatest distance occurs when theta = 0 degrees, which means the line connecting the sun and the point on the ellipse is aligned with the major axis of the ellipse.
Plugging in the values, we get:
r_max = a(1 - e^2) / (1 + e*cos(0°))
Simplifying the equation gives:
r_max = a(1 - e^2) / (1 + e)
Similarly, the least distance occurs when theta = 180 degrees, which means the line connecting the sun and the point on the ellipse is aligned with the major axis, but in the opposite direction.
Plugging in the values, we get:
r_min = a(1 - e^2) / (1 + e*cos(180°))
Simplifying the equation gives:
r_min = a(1 - e^2) / (1 - e)
By substituting the given values for a and e, you can calculate the greatest and least distances of the earth from the sun.
2. To solve for the angle that the battleship should direct its missile with respect to point A in order to hit point B, we can use trigonometry and the given information.
Let's break down the problem step by step:
- The battleship is traveling east at 40 km/h, so after 2 hours, it would have traveled a distance of 40 km/h * 2 hours = 80 km. This puts the battleship at point C.
- The smaller ship is traveling north at 60 km/h, so after 2 hours, it would have traveled a distance of 60 km/h * 2 hours = 120 km. This puts the smaller ship at point A.
- The distance between point A and point B is given as 150 km.
- The angle between the direction of the battleship (east) and the line connecting point A and point But (10 degrees north of east) is what we need to find.
To solve for the desired angle, we can use trigonometry. Since we have a right triangle formed by points A, B, and But, we can use the tangent function to find the angle.
Tangent of an angle = opposite / adjacent
In this case, the opposite side is the distance between points A and But (150 km) and the adjacent side is the distance between points But and B.
Let's call the angle we want to find theta. We have:
tan(theta) = opposite / adjacent
tan(theta) = 150 km / adjacent
To find the adjacent side, we need to use the Pythagorean theorem, which states that:
(c^2) = (a^2) + (b^2)
In this case, the adjacent side is the hypotenuse of a right triangle formed by points But, B, and C. So, we have:
adjacent^2 = (80 km)^2 + (h)^2
Now we can substitute this into the equation for the tangent:
tan(theta) = 150 km / sqrt((80 km)^2 + (h)^2)
Solving for theta, we can take the inverse tangent (arctan) of both sides:
theta = arctan(150 km / sqrt((80 km)^2 + (h)^2))
By substituting the given values for the distance between points A and But (150 km), and using the equation for the adjacent side, h, you can calculate the angle theta.