A plane leaves airport A and travels 570 miles to airport B on a bearing of Upper N 32 degrees Upper E. The plane later leaves a
irport B and travels to airport C 390 miles away on a bearing of Upper S 73 degrees Upper E. Find the distance from airport A to airport C to the nearest tenth of a mile.
1 answer:
Answer:
769.46 miles
Step-by-step explanation:
Please have a look at the photo attached in this answer
The missing part of the angle is 32 degrees, so the total angle of B is:
32 + 73 = 105 degrees.
Let x is the distance from airport A to airport C to the nearest tenth of a mile, we use cosine law to find x:
=
- 2(570)(390)cos(105)
<=> = 592070
<=> x = = 769.46 miles
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Given <span>the paraboloid
We can rewrite it as a scalar value function f as follows:
The normal to the </span><span>paraboloid at any point is given by:
Also, the normal to the given plane
Equating the two normal vectors, we have:
</span>
Since, -1 = 2 is not possible, therefore there exist no such point <span>on the paraboloid