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A 10-foot board rests against a wall. The angle that the board makes with the ground is [tex]$60^{\circ}$[/tex]. How far is the base of the board away from the wall?

Select the correct trig ratio and distance from the wall:

A. [tex]\cos 60^{\circ}=\frac{10}{x} ; x=20[/tex] feet

B. [tex]\sin 60^{\circ}=\frac{10}{x} ; x \approx 11.5[/tex] feet

C. [tex]\cos 60^{\circ}=\frac{x}{10} ; x=5[/tex] feet

D. [tex]\sin 60^{\circ}=\frac{x}{10} ; x \approx 8.66[/tex] feet



Answer :

GinnyAnswer
To solve this problem, we need to use the appropriate trigonometric ratio based on the given angle and the length of the board.

Given:
- The length of the board \( AB \) is 10 feet.
- The angle \( \theta \) between the board and the ground is \( 60^\circ \).

We need to find the horizontal distance \( x \) from the base of the board to the wall (point \( A \) to the wall at point \( C \)). This scenario forms a right triangle \( ABC \) where:
- \( AC \) (the distance we need to find) is the adjacent side of the angle \( 60^\circ \).
- \( AB \) (the length of the board) is the hypotenuse.

The cosine function relates the adjacent side to the hypotenuse in a right triangle. Specifically,
[tex]\[ \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \][/tex]

Using the given values:
[tex]\[ \cos(60^\circ) = \frac{x}{10} \][/tex]

We know that \( \cos(60^\circ) \) is \( \frac{1}{2} \). Hence,
[tex]\[ \frac{1}{2} = \frac{x}{10} \][/tex]

To solve for \( x \), we multiply both sides by 10:
[tex]\[ x = 10 \times \frac{1}{2} \][/tex]
[tex]\[ x = 5 \][/tex]

Therefore, the base of the board (point \( A \)) is 5 feet away from the wall (point \( C \)).

The correct trig ratio and distance from the wall, according to the choices given, is:

C. [tex]\(\cos 60^\circ = \frac{x}{10} ; x=5\)[/tex] feet

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