Full solution

Q. The area of a parallelogram is

10

, and the lengths of its sides are

6

9

and

6

7

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

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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 also be expressed as $h = b \times \sin(\theta)$ , where $\theta$ is the angle between the base and the side adjacent to the base.
Choosing Base: We are given the area $A = 10$ and the lengths of the sides, but we do not know which side should be considered the base. Since the formula for the area does not change with the choice of base and height as long as they are perpendicular, we can choose either side as the base. Let's choose the longer side, $6.9$ , as the base for convenience.
Height Calculation: Now we can express the height $h$ in terms of the angle $\theta$ and the base $b$ . We have $h = 6.9 \times \sin(\theta)$ . We can substitute this into the area formula to get $10 = 6.9 \times h = 6.9 \times 6.9 \times \sin(\theta)$ .
Solving for $\sin(\theta)$ : To find $\sin(\theta)$ , we rearrange the equation to solve for $\sin(\theta)$ : $\sin(\theta) = \frac{10}{(6.9 \times 6.9)}$ .
Calculating $\sin(\theta)$ : Calculating $\sin(\theta)$ , we get $\sin(\theta) = \frac{10}{(6.9 \times 6.9)} = \frac{10}{47.61} \approx 0.210$ .
Finding Angle $\theta$ : Now we need to find the angle $\theta$ whose sine is approximately $0.210$ . We use the inverse sine function (also known as arcsin) to find this angle. $\theta = \arcsin(0.210)$ .
Final Angle Calculation: Using a calculator, we find that $\theta \approx \arcsin(0.210) \approx 12.1$ degrees. Since we are looking for the acute angle and the arcsin function returns the acute angle, this is the angle we are looking for.