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solve for ( x ). round to the nearest tenth, if necessary. right triang…

Question

solve for ( x ). round to the nearest tenth, if necessary.
right triangle with right angle at v, side vw labeled ( x ), side wu (hypotenuse) labeled 2.6, angle at w is ( 36^circ )
answer
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Explanation:

Step1: Identify adjacent side, hypotenuse

In right $\triangle VWU$, $\angle W=36^\circ$, hypotenuse $WU=2.6$, side $x$ (VW) is adjacent to $\angle W$.

Step2: Use cosine ratio

Cosine of angle = $\frac{\text{adjacent}}{\text{hypotenuse}}$
$\cos(36^\circ) = \frac{x}{2.6}$

Step3: Solve for $x$

$x = 2.6 \times \cos(36^\circ)$
Calculate $\cos(36^\circ) \approx 0.8090$, so $x \approx 2.6 \times 0.8090$
$x \approx 2.1034$

Step4: Round to nearest tenth

$x \approx 2.1$

Answer:

$x = 2.1$