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The area of a parallelogram is 1033 , and the lengths of its sides are 67 and 32 . Determine, to the nearest tenth of a degree, the measure of the acute angle of the parallelogram. Answer:

The area of a parallelogram is 10331033 , and the lengths of its sides are 6767 and 3232 . Determine, to the nearest tenth of a degree, the measure of the acute angle of the parallelogram.\newlineAnswer:

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Q. The area of a parallelogram is 10331033 , and the lengths of its sides are 6767 and 3232 . Determine, to the nearest tenth of a degree, the measure of the acute angle of the parallelogram.\newlineAnswer:
  1. Identify Formula: To find the acute angle of the parallelogram, we need to use the formula for the area of a parallelogram, which is A=base×heightA = \text{base} \times \text{height}. Here, the base can be either side of the parallelogram, and the height is the perpendicular distance from the base to the opposite side. We can consider one of the given side lengths as the base and the other as the product of the other side and the sine of the acute angle between them.
  2. Choose Base and Height: Let's choose the longer side, 6767, as the base. Then the height can be represented as 32×sin⁡(θ)32 \times \sin(\theta), where θ\theta is the acute angle we are trying to find. The area of the parallelogram is given as 10331033, so we can set up the equation 67×32×sin⁡(θ)=103367 \times 32 \times \sin(\theta) = 1033.
  3. Set Up Equation: Now, we solve for sin⁡(θ)\sin(\theta). Dividing both sides of the equation by (67×32)(67 \times 32), we get sin⁡(θ)=1033(67×32)\sin(\theta) = \frac{1033}{(67 \times 32)}.
  4. Solve for sin⁡(θ)\sin(\theta): Calculating the right side of the equation, we have sin⁡(θ)=1033(67×32)=10332144≈0.4817\sin(\theta) = \frac{1033}{(67 \times 32)} = \frac{1033}{2144} \approx 0.4817.
  5. Calculate sin⁡(θ):\sin(\theta): To find the angle θ\theta, we take the inverse sine (arcsin) of 0.48170.4817. Using a calculator, we find θ≈arcsin⁡(0.4817)\theta \approx \arcsin(0.4817).
  6. Find Angle θ\theta: Calculating the arcsin⁡(0.4817)\arcsin(0.4817), we get θ≈28.9\theta \approx 28.9 degrees to the nearest tenth of a degree.

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