I need some help with this question
1 answer:
You might be interested in
Answer:
Step-by-step explanation:
We have plane 1 flying SW for 4 hours at a rate of 415 mph. The distance he covers using the d = rt formula for distance, is 415(4) = 160 miles.
We also have plane 2 flying directly east (along the x-axis) for 4 hours at 327 mph. The distance he covers using the d = rt formula for distance, is 327(4) = 1308 miles. The angle in between them at this point is 135 degrees, and what we need to find is the length of the vector connecting the 2 planes. IF this was right triangle trig that distance would be the hypotenuse and we could solve for it using Pythagorean's Theorem. BUT it is NOT a right triangle, so we have to find some other means with which to solve for that length. We will use the Law of Cosines to do this.
which simplifies a bit to
If you add all of that together, you'll get
and you'll take the square root of that to get that the distance between the 2 planes after 4 hours is
2745 miles
Answer:
a) Acute triangle
b) Right triangle
Step-by-step explanation:
∵ A triangle having sides a, b and c is called,
Acute : If a² + b² > c² or a² + c² > b² or b² + c² > a²,
Obtuse : if a² + b² < c² or a² + c² < b² or b² + c² < a²
Right : a² + b² = c² or a² + c² = b² or b² + c² = a²,
a) Let A≡(-1, 1), B≡(7,-2) and C≡(1,-5),
By the distance formula,
Similarly,
∵ The sum of any two sides is greater than third side,
So, ABC is a triangle,
Now,
⇒ ABC is an acute triangle.
b) Let P≡(0,6), Q≡(1,2) and R≡(5,3),
By the distance formula,
PQ = √17 unit,
QR = √17 unit,
RP = √34 unit,
∵ The sum of any two sides of PQR is greater than third side,
⇒ PQR is a triangle ,
Also,
Hence, PQR is a right triangle.