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A phane takes off at an angle of elevation of 15 ° and travels in a straight line for 3,000 meters. What is the height of plane above the ground at this instant ?

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3 votes

Final answer:

The height of the plane above the ground after traveling 3,000 meters at a 15° angle of elevation is approximately 776.4 meters. This is calculated using the sine function: height = 3,000 m × sin(15°).

Step-by-step explanation:

To find the height of the plane above the ground after taking off at an angle of elevation of 15° and traveling for 3,000 meters, we can use trigonometric functions. Specifically, the sine function relates the opposite side of a right-angled triangle (in this case, the height above ground) to the hypotenuse (the distance traveled by the plane).

The sine of an angle in a right triangle is equal to the length of the opposite side divided by the length of the hypotenuse. The formula is:

height = hypotenuse × sin(angle)

Here, the hypotenuse is the distance traveled by the plane (3,000 m), and the angle is 15°. Thus, the calculation would be:

height = 3,000 m × sin(15°)

We can calculate the sine of 15° using a calculator.

height = 3,000 m × 0.2588 (approximately)

height = 776.4 m (approximately)

Therefore, the plane is approximately 776.4 meters above the ground.

User Felixqk
by
7.7k points
4 votes
Using SOHCAHTOA we have:
sin15 = height / 3000
height = 3000 sin 15 height = 776.46m

So, the height of the plane above the ground at this instant is 776.46m
User Srichand Yella
by
9.1k points

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