Question

solve the triangle : BC=1140in, AC=854in, AB=771in

Answer 1

Question#20645

solve the triangle: BC=1140in, AC=854in, AB=771in

Solution

Using the law of cosines find the angle A:

$$BC^2 = AC^2 + AB^2 - 2 \cdot AC \cdot AB \cdot \cos(A);$$ $$\cos(A) = \frac{AC^2 + AB^2 - BC}{2 \cdot AC \cdot AB};$$ $$\cos(A) = \frac{854^2 + 771^2 - 1140^2}{2 \cdot 854 \cdot 771} = 0,018344;$$ $$A = 88,95{}^\circ;$$

Next using the law of sines:

$$\frac{BC}{\sin(A)} = \frac{AC}{\sin(B)};$$ $$\sin(B) = \frac{AC \cdot \sin(A)}{BC};$$ $$\sin(B) = \frac{854 \cdot 0,9998}{1140} = 0,748997;$$ $$B = 48,5{}^\circ$$

And finally the rule of angels of triangles gives us:

$$A + B + C = 180{}^\circ;$$ $$C = 180{}^\circ - A - B;$$ $$C = 32,55{}^\circ;$$

Answer:

$$A = 88,95{}^\circ; B = 48,5{}^\circ; C = 32,55{}^\circ;$$