Sketch a diagram describing the given situation, and then recall that the distance is a product of velocity and time.
5 mi
Practice makes perfect
We are given that an airplane at a constant altitude a flies a horizontal distance d toward us at velocity v, and observing for time t we measure angles of elevation ∠ E_1 and ∠ E_2 at the start and end of our observation. Let's sketch a diagram describing this situation.
Now let's substitute the values we are given in our exercise. Notice that since ∠ E_2 is 90^(∘), the airplane is exactly above our head at the end.
Since we are given that the velocity of this airplane is 5 miles per minute and we observe the plane for 1 minute, we can evaluate the distance d. Let's recall that the distance is the product of velocity and time.
d=vt ⇒ d=5*1=5
The distance this airplane traveled is 5 miles. Let's add this information to our diagram.
To find the altitude a, we will use one of the trigonometric ratios. Let's recall that the tangent of an angle is the ratio between the side opposite to this angle and the side adjacent to this angle. Using this definition, we can write an equation for tan 45^(∘).
tan 45^(∘)=a/5
Let's solve the above equation.
