Two planes leave an airport at the same time, one flying at 300 km/h and the other at 420 km/h The angle between their flight paths is 75 degrees. Aftar three hours, how far apart are they ?
i guss i have to use the equation D= b.h but i cant fit the number in spot can u please help me
This is a vector problem. Let the x axis be the direction that the 300 km/h plane flies. Assume the faster plane flies at 75 degrees to that direction with a velocity component along both the +x and +y axes.
The location of plane 1 at time t is given by
X1 = 300 t
Y1 = 0
The location of plane 2 at time 2 is
X2 = 420 cos 75 t = 108.7 t
Y2 = 420 sin 75 t = 405.7 t
The distance between than at any time t is
sqrt [(X2-X1)^2 + (Y2-Y1)^2]
Plug in t = 3 hours and solve
I n my previous answer, I solved it using a vector method, not a trig method. Both will give the same answer. To use trigonometry, draw a triangle with two sides representing the two distances and directions of the planes after three hours. The angle between these two sides is 75 degrees. The third side of the triangle is the line between the planes at that time.
You have two sides of a triangle and the included angle (75 degrees). The law of cosines can be used for the third side, which is what you want.
please show me again in trig way
To solve this problem, we can use the concept of vectors and the law of cosines. Let's break it down step by step:
Step 1: Understand the problem.
Two planes leave an airport at the same time, one flying at 300 km/h and the other at 420 km/h. We need to find out how far apart they are after three hours.
Step 2: Visualize the problem.
To visualize the problem, we can draw a diagram. Let's assume that the two planes start at point A (the airport). Plane 1 flies at 300 km/h along path AB, and Plane 2 flies at 420 km/h along path AC. The angle between the flight paths AB and AC is given as 75 degrees. After three hours, we want to determine the distance between points B and C.
B
/|
300 / | 420
/ |
/ |
A/_____|C
Step 3: Apply the law of cosines.
The law of cosines states that in a triangle, the square of one side equals the sum of the squares of the other two sides minus twice the product of the lengths of those two sides multiplied by the cosine of the included angle.
In this case, we can apply the law of cosines to find the side BC (distance between planes after three hours).
The formula for the law of cosines is:
c^2 = a^2 + b^2 - 2ab * cos(C)
Let's assign the sides of the triangle as follows:
a = 300 km/h (the speed of Plane 1)
b = 420 km/h (the speed of Plane 2)
C = 75 degrees
Substituting the given values into the formula, we get:
BC^2 = (300 km/h)^2 + (420 km/h)^2 - 2 * (300 km/h) * (420 km/h) * cos(75 degrees)
Step 4: Calculate the distance.
Using a calculator or mathematical software, we can now calculate the value of BC^2.
BC^2 = (90000 km^2/h^2) + (176400 km^2/h^2) - 2 * (126000 km^2/h^2) * cos(75 degrees)
BC^2 = 266400 km^2/h^2 - 252000 km^2/h^2 * cos(75 degrees)
BC^2 = 266400 km^2/h^2 - 239561.5304 km^2/h^2
BC^2 = 26838.4696 km^2/h^2
Taking the square root of both sides, we find the value of BC:
BC = sqrt(26838.4696 km^2/h^2)
BC ≈ 163.94 km/h
After three hours, the two planes are approximately 163.94 km apart.