Question

A weather balloon is launched from the ground at a spot 250 yards from a 65 foot royal palm tree . What is the minimal angle of elevation at which the balloon must take off in order to avoid hitting the tree? Assume that the balloon flies in a straight line and the angle of elevation stays constant.

Answer 1

Answer:

Angle of elevation at which the balloon must take off in order to avoid hitting the tree is 5 degrees

Step-by-step explanation:

Given:

The height of the tree  = 65 foot

The distance between the tree and the spot from which the balloon is launched = 250 yards

To find:

The angle of elevation at which the balloon must take off in order to avoid hitting the tree = ?

Solution:

Converting the yards to feet

1 yards = 3 feet

250 yards = 3 x 250 = 750 yards

Refer the below figure, The angle x is the angle of elevation at which the the ball must be thrown so that it does not Hit the tree

$tan(x) = \frac{opposite}{adjacent}$

Substituting the values

$tan(x) = \frac{65}{750}$

$tan(x) = 0.086$

$x = tan^{-1}(0.086)}$

$x = 4.91^{\circ}$

$x \approx5^{\circ}$

Answer 2

Answer:

4.95 (not rounded)

Step-by-step explanation:

1 yard = 3 foot

250 yards X 3 foot =750 foot

Tanθ 65/750

θ= tan -1 (65/750)

=4.95