Question

solve for ( x ). round to the nearest tenth, if necessary.
(there is a right triangle pon with right angle at o, angle at p is ( 60^circ ), hypotenuse pn is 9, and side po is ( x ))
answer attempt 1 out of 2
( x = ) input box submit answer

Explanation:

Step1: Identify the trigonometric ratio

In right triangle \( PON \), we have angle \( P = 60^\circ \), hypotenuse \( PN = 9 \), and adjacent side to angle \( P \) is \( x \) (since \( \angle O \) is the right angle, \( PO \) is adjacent to \( \angle P \)). The cosine of an angle in a right triangle is defined as the adjacent side over the hypotenuse. So, \( \cos(60^\circ)=\frac{x}{9} \).

Step2: Solve for \( x \)

We know that \( \cos(60^\circ) = 0.5 \). Substituting this value into the equation from Step 1, we get:
\( 0.5=\frac{x}{9} \)
To solve for \( x \), multiply both sides of the equation by 9:
\( x = 9\times0.5 \)
\( x = 4.5 \)

Answer:

\( 4.5 \)