Question - Finding Included Angle of a Triangle
Solution:
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 $$43.6^{\circ}$$.
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