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The area of a triangle is 6 . Two of the side lengths are 7.1 and 5.3 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 66 . Two of the side lengths are 77.11 and 55.33 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 66 . Two of the side lengths are 77.11 and 55.33 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, we can use the formula for the area of a triangle, which is given by A=12absin⁡(C) A = \frac{1}{2}ab\sin(C) , where A A is the area, a a and b b are the lengths of two sides, and C C is the included angle between those sides.\newlineGiven: \newlineArea A=6 A = 6 square units,\newlineSide a=7.1 a = 7.1 units,\newlineSide b=5.3 b = 5.3 units.\newlineWe need to solve for C C .
  2. Calculate Sin(C): First, we rearrange the area formula to solve for sin⁡(C) \sin(C) :\newlinesin⁡(C)=2Aab \sin(C) = \frac{2A}{ab} .\newlineSubstitute the given values into the formula:\newlinesin⁡(C)=2×67.1×5.3 \sin(C) = \frac{2 \times 6}{7.1 \times 5.3} .
  3. Find Angle C: Now, we calculate the value of sin⁡(C) \sin(C) :\newlinesin⁡(C)=127.1×5.3 \sin(C) = \frac{12}{7.1 \times 5.3} .\newlinesin⁡(C)=1237.63 \sin(C) = \frac{12}{37.63} .\newlinesin⁡(C)≈0.3189 \sin(C) \approx 0.3189 .
  4. Calculate Angle C: Next, we find the angle C C by taking the inverse sine (arcsin) of sin⁡(C) \sin(C) :\newlineC=arcsin⁡(0.3189) C = \arcsin(0.3189) .
  5. Calculate Angle C: Next, we find the angle C C by taking the inverse sine (arcsin) of sin⁡(C) \sin(C) :\newlineC=arcsin⁡(0.3189) C = \arcsin(0.3189) .Using a calculator, we find that:\newlineC≈arcsin⁡(0.3189) C \approx \arcsin(0.3189) .\newlineC≈18.8 C \approx 18.8 degrees (to the nearest tenth of a degree).

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