Answer: 42.21 km
Step-by-step explanation:
We can solve this using trigonometry, since we have the following data:
is the the angle of elevation
is the horizontal distance between the plane and the radar station
is the hypotenuse of the right triangle formed between the radar station and the airplane
Now, the trigonometric function that will be used is <u>cosine</u>:
because 
is the adjacent side of the right triangle
Finding :
Answer:
y = 140, x = 159
Step-by-step explanation:
Angle y:
we can find this angle by subtracting 40 from 180, 180-40 = 140 = y
Angle x:
Some measure of an angle plus angle b is equal to 180 degrees: 180-61 = 119
We can now find the measure of the third angle of the triangle: 180-(119+40) = 21
The third angle plus angle x is equal to 180: 180-21 = 159 = x
Step-by-step explanation:
based on your equation it would look like this. use Dennis its a life saver
In order to do this, you must first find the "cross product" of these vectors. To do that, we can use several methods. To simplify this first, I suggest you compute:
‹1, -1, 1› × ‹0, 1, 1›
You are interested in vectors orthogonal to the originals, which don't change when you scale them. Using 0,-1,1 is much easier than 6s and 7s.
So what methods are there to compute this? You can review them here (or presumably in your class notes or textbook):
http://en.wikipedia.org/wiki/Cross_produ...
In addition to these methods, sometimes I like to set up:
‹1, -1, 1› • ‹a, b, c› = 0
‹0, 1, 1› • ‹a, b, c› = 0
That is the dot product, and having these dot products equal zero guarantees orthogonality. You can convert that to:
a - b + c = 0
b + c = 0
This is two equations, three unknowns, so you can solve it with one free parameter:
b = -c
a = c - b = -2c
The computation, regardless of method, yields:
‹1, -1, 1› × ‹0, 1, 1› = ‹-2, -1, 1›
The above method, solving equations, works because you'd just plug in c=1 to obtain this solution. However, it is not a unit vector. There will always be two unit vectors (if you find one, then its negative will be the other of course). To find the unit vector, we need to find the magnitude of our vector:
|| ‹-2, -1, 1› || = √( (-2)² + (-1)² + (1)² ) = √( 4 + 1 + 1 ) = √6
Then we divide that vector by its magnitude to yield one solution:
‹ -2/√6 , -1/√6 , 1/√6 ›
And take the negative for the other:
‹ 2/√6 , 1/√6 , -1/√6 ›
