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The area of a parallelogram is 18 , and the lengths of its sides are 9.2 and 2 . Determine, to the nearest tenth of a degree, the measure of the acute angle of the parallelogram.
Answer:

The area of a parallelogram is $18$ , and the lengths of its sides are $9$ . $2$ and $2$ . Determine, to the nearest tenth of a degree, the measure of the acute angle of the parallelogram. $\newline$ Answer:

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Find the magnitude of a three-dimensional vector

Full solution

Q. The area of a parallelogram is

18

, and the lengths of its sides are

9

.

2

and

2

. Determine, to the nearest tenth of a degree, the measure of the acute angle of the parallelogram.

\newline

Answer:

Area Formula: The area of a parallelogram is given by the formula $A = b \times h$ , where $A$ is the area, $b$ is the base, and $h$ is the height. The height is the perpendicular distance from the base to the opposite side, which can be expressed as $h = \frac{A}{b}$ when the area and base are known.
Calculate Height: We know the area $A = 18$ and one side length $b = 9.2$ . We can calculate the height $h$ using the formula $h = \frac{A}{b}$ . $\newline$ $h = \frac{18}{9.2}$ $\newline$ $h = 1.95652173913$ (approximately)
Calculate Angle: The height we have found is the side length multiplied by the sine of the acute angle, since $h = b \times \sin(\theta)$ . We can rearrange this to find the angle $\theta$ : $\theta = \arcsin(\frac{h}{b})$ .
Final Angle Calculation: We substitute the values we have into the formula to find $\theta$ . $\theta = \arcsin(\frac{1.95652173913}{9.2})$ $\theta = \arcsin(0.21267434771)$ $\theta = 12.3^\circ$ (approximately, to the nearest tenth of a degree)

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