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The area of a triangle is 3 . Two of the side lengths are 1.8 and 4.8 and the included angle is acute. Find the measure of the included angle, to the nearest tenth of a degree. Answer:

The area of a triangle is 33 . Two of the side lengths are 11.88 and 44.88 and the included angle is acute. Find the measure of the included angle, to the nearest tenth of a degree.\newlineAnswer:

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Q. The area of a triangle is 33 . Two of the side lengths are 11.88 and 44.88 and the included angle is acute. Find the measure of the included angle, to the nearest tenth of a degree.\newlineAnswer:
  1. Area Formula: To find the measure of the included angle in a triangle given its area and two side lengths, we can use the formula for the area of a triangle, which is A=12absin⁡(C) A = \frac{1}{2}ab\sin(C) , where A A is the area, a a and b b are the side lengths, and C C is the included angle.
  2. Calculate Product: We know the area A=3 A = 3 , side length a=1.8 a = 1.8 , and side length b=4.8 b = 4.8 . We can plug these values into the area formula to solve for sin⁡(C) \sin(C) .\newline3=12×1.8×4.8×sin⁡(C) 3 = \frac{1}{2} \times 1.8 \times 4.8 \times \sin(C)
  3. Divide Area: First, calculate the product of 12 \frac{1}{2} , 1.8 1.8 , and 4.8 4.8 .\newline12×1.8×4.8=4.32 \frac{1}{2} \times 1.8 \times 4.8 = 4.32
  4. Find Sin(C): Now, divide the area by this product to solve for sin⁡(C) \sin(C) .\newlinesin⁡(C)=34.32 \sin(C) = \frac{3}{4.32}
  5. Inverse Sine: Perform the division to find sin⁡(C) \sin(C) .\newlinesin⁡(C)=0.6944... \sin(C) = 0.6944...
  6. Calculate Angle: To find the angle C C , we need to take the inverse sine (arcsin) of sin⁡(C) \sin(C) .\newlineC=arcsin⁡(0.6944...) C = \arcsin(0.6944...)
  7. Calculate Angle: To find the angle C C , we need to take the inverse sine (arcsin) of sin⁡(C) \sin(C) .\newlineC=arcsin⁡(0.6944...) C = \arcsin(0.6944...) Using a calculator, we find the value of C C to the nearest tenth of a degree.\newlineC≈43.9∘ C \approx 43.9^\circ

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