Question

A ball is dropped from the top of a 45 foot building. Once the ball is released, a gust of wind blows the ball off course and is dropped 4 feet from the base of the building. How many degrees was the ball thrown off course?

Answer 1

the ball thrown off course was  $x = 5$° .

Step-by-step explanation:

Here we have , A ball is dropped from the top of a 45 foot building. Once the ball is released, a gust of wind blows the ball off course and is dropped 4 feet from the base of the building. We need to find that How many degrees was the ball thrown off course? Let's find out:

Basically above given condition is a case of right angled triangle with following parameters

$Perpendicular = 4 ft$

$Base = 45 ft$

x = angle between Perpendicular & base

Now , $tanx = \frac{Perpendicular}{base}$

⇒ $tanx = \frac{Perpendicular}{base}$

⇒ $tanx = \frac{4}{45}$

Taking tan inverse both sides , in order to obtain x :

⇒ $tan^{-1}(tanx) = tan^{-1}(\frac{4}{45})$

⇒ $x = tan^{-1}(\frac{4}{45})$

⇒ $x = tan^{-1}(0.08)$

⇒ $x = 5$°

Therefore, the ball thrown off course was  $x = 5$° .

Answer 2

Answer:

The degrees the ball thrown off course is given by 5.08° degrees.

Step-by-step explanation:

The triangle uploaded in the figure shows the current scenario has the distance from the top to bottom is 45 feet, and the distance blown due to the wind is 4 feet.

The figure shown in the triangle is a right-angled triangle. Hence we can apply Pythagoras' theorem.

Hence hypotenuse side length = √(45²+4²)

= 45.18

from the triangle , tan(A) =45÷4

A = 84.92°

B = 180-90-84.92

= 5.08°

Therefore, the degrees the ball thrown off course is given by 5.08° degrees.