How do the height and base length of the two triangles compare to the height and base length of the original piece of paper?
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Answer:
40.88 miles
Step-by-step explanation:
Here I attach a draw of the situation.
Our plane is flying at 20,000 ft and we want it to be at 35,000 ft, thus, we need to take it up 15,000 ft high. We will do so with an elevation angle of 4dg. Notice what we have is a triangle, more precisely a right triangle, which angles are 90dg, 4dg and, consequently, 86dg. Also, we know that one of the legs is 15,000. The other legs we do not know yet and neither the hypotenuse. However, we can find this measures using trigonometry.
We know that, given an angle a, it mus be that:
sin(a) = opposite leg / hypotenuse
Take a ad our angle of 4dg. So, sin(a)=sin(4)= 0.070
And the opposite leg to a measures 15000ft. Thus,
0.070 = 15000/hypotenuse
Dividing both sides by 0.070:
1 = 15000/0.070 * hypotenuse
Multiplying both sides by hypotenuse:
hypotenuse = 15000/0.070
hypotenuse = 215,033.9
So, our plane will need to fly 215,033.9 ft to get to the height of 35000 ft.
However, we need the result in miles. As 1 mile is 5260 ft, we just need to divide our result in ft by 5260:
215,033.9/5260 = 40.88 miles
So, the plane need to fly 40.88 miles at an elevation angle of 4dg to reach a height of 35,000 ft.
