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ratelena [41]
3 years ago
13

Speed of Plane A = 425 mph. Speed of Plane B = 375 mph. Both planes leave from Kansas City at the same time. Plane A flies due W

est. Plane B flies due South. After 3 hours, how far is Plane A from Plane B?
Mathematics
1 answer:
Anna007 [38]3 years ago
3 0
After 3 hours, Plane A has flown (425 mi/h)*(3 h) = 1275 mi.
After 3 hours, Plane B has flown (375 mi/h)*(3 h) = 1125 mi.

Since the distances flown are in directions that are at a right angle to each other, the Pythagorean theorem can be used to figure the separation. The distance between planes after 3 hours is
  √(1275² +1125²) = √2,891,250 ≈ 1700.4

After 3 hours, Plane A is about 1700 miles from Plane B.
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Answer:

The measure of the side AB is 26 cm.

Step-by-step explanation:

The question is:

In a triangle ABC the hypotenuse BC measures 30 cm, it is known that the angle of the opposite angle to the side AB, measures 60°, therefore it is concluded that as side AB the triangle measures ?

Solution:

Consider the triangle ABC.

The side BC is defined as a hypotenuse. This implies that the triangle ABC is a right angled triangle.

The angle A measures 90° and the angle C measures 60°.

The hypotenuse length is 30 cm.

According to the trigonometric identities for right angled triangle:

sin\ \theta^{o}=\frac{Perpendicular}{Hypotenuse}

Compute the length of side AB as follows:

sin\ \theta^{o}=\frac{Perpendicular}{Hypotenuse}

sin\ 60^{o}=\frac{AB}{BC}\\\\0.866=\frac{AB}{30}\\\\AB=30\times 0.866\\\\AB=25.98\\\\AB\approx 26

Thus, the measure of the side AB is 26 cm.

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