Question

solve for x. round to the nearest tenth, if necessary.

Explanation:

Step1: Identify trigonometric ratio

In right triangle \( \triangle UVT \), \( \angle T = 70^\circ \), \( UT = 64 \) (adjacent to \( \angle T \)), and \( UV = x \) (opposite to \( \angle T \))? Wait, no—wait, \( \angle U \) is right angle, so \( \angle T = 70^\circ \), \( UT = 64 \) (adjacent to \( \angle T \)), \( UV = x \) (opposite to \( \angle T \))? Wait, no, \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \). Wait, \( \angle T = 70^\circ \), so opposite side to \( \angle T \) is \( UV = x \), adjacent is \( UT = 64 \). So \( \tan(70^\circ) = \frac{x}{64} \)? Wait, no—wait, \( \angle V \) would be \( 20^\circ \), but maybe better to use \( \tan(70^\circ) = \frac{UV}{UT} \)? Wait, no, \( \angle T = 70^\circ \), right angle at \( U \), so sides: \( UT = 64 \) (leg), \( UV = x \) (leg), \( VT \) hypotenuse. Wait, \( \tan(\angle T) = \frac{UV}{UT} \), so \( \tan(70^\circ) = \frac{x}{64} \)? Wait, no, \( \angle T = 70^\circ \), so the side opposite \( \angle T \) is \( UV \) (length \( x \)), adjacent is \( UT \) (length 64). So \( \tan(70^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{64} \). Then solve for \( x \): \( x = 64 \times \tan(70^\circ) \).

Step2: Calculate \( \tan(70^\circ) \)

\( \tan(70^\circ) \approx 2.7475 \)

Step3: Compute \( x \)

\( x = 64 \times 2.7475 \approx 175.84 \). Wait, that seems large. Wait, maybe I mixed up opposite and adjacent. Wait, \( \angle T = 70^\circ \), right angle at \( U \), so \( UT \) is adjacent to \( \angle T \), \( UV \) is opposite? Wait, no—wait, \( \angle T \) is at vertex \( T \), so the sides: \( TU \) is one leg (length 64), \( UV \) is the other leg (length \( x \)), \( VT \) is hypotenuse. So angle at \( T \): between \( TU \) (adjacent) and \( VT \) (hypotenuse), and opposite side is \( UV \). So \( \tan(\angle T) = \frac{UV}{TU} \), so \( \tan(70^\circ) = \frac{x}{64} \), so \( x = 64 \tan(70^\circ) \approx 64 \times 2.7475 \approx 175.8 \). Wait, but maybe it's \( \cot(70^\circ) \)? Wait, no—wait, maybe I got the angle wrong. Wait, angle at \( V \): since it's a right triangle, angles sum to \( 180^\circ \), so \( \angle V = 180 - 90 - 70 = 20^\circ \). Then, if we take \( \angle V = 20^\circ \), then \( \tan(20^\circ) = \frac{UT}{UV} = \frac{64}{x} \), so \( x = \frac{64}{\tan(20^\circ)} \). \( \tan(20^\circ) \approx 0.3640 \), so \( x = \frac{64}{0.3640} \approx 175.8 \). Same result. So that's correct.

Answer:

\( \boxed{175.8} \)