Finding Included Angle of a Triangle
The question in the image is regarding finding the measure of the included angle of a triangle, given the area and the lengths of two sides. The formula to find the area of a triangle when two sides and the included angle are known is:
\[ \text{Area} = \frac{1}{2} ab \sin(C) \]
Where:
- \(a\) and \(b\) are the lengths of the two sides,
- \(C\) is the included angle,
- \( \text{Area} \) is the area of the triangle.
Given:
- \(a = 55 \),
- \(b = 52 \),
- \( \text{Area} = 986 \).
Let's find the angle \(C\):
\[ 986 = \frac{1}{2} \times 55 \times 52 \times \sin(C) \]
\[ 986 = \frac{1}{2} \times 2860 \times \sin(C) \]
\[ \sin(C) = \frac{986}{1430} \]
\[ \sin(C) = 0.68951... \]
Now, take the inverse sine (arc sine) of both sides to find the angle \(C\):
\[ C = \arcsin(0.68951...) \]
Using a calculator, you'll find that:
\[ C \approx 43.60282^{\circ} \]
Therefore, the measure of the included angle, rounded to the nearest tenth of a degree, is approximately \(43.6^{\circ}\).
