Solve $\triangle A B C$ subject to the given conditions, if possible. Round the lengths of the sides and measures of the angles (in degrees) to 1 decimal place if necessary. Round intermediate steps to at least four decimal places.
b=600, a=500, C=133°
$c \approx$
Answer
Rounding to one decimal place, we get the final answer: \(c \approx \boxed{1009.6}\).
Step 1 ：We are given a triangle ABC with side lengths a, b and angle C. The given values are a=500, b=600, and C=133°.
Step 2 ：We can use the Law of Cosines to find the length of side c. The Law of Cosines states that for any triangle with sides of lengths a, b, and c, and an angle C opposite side c, the following equation holds: \(c² = a² + b² - 2ab\cos(C)\).
Step 3 ：We can rearrange this equation to solve for c: \(c = \sqrt{a² + b² - 2ab\cos(C)}\).
Step 4 ：Substituting the given values into the equation, we get \(c = \sqrt{500² + 600² - 2*500*600\cos(133°)}\).
Step 5 ：Calculating the above expression, we find that \(c \approx 1009.5538698046277\).
Step 6 ：Rounding to one decimal place, we get the final answer: \(c \approx \boxed{1009.6}\).
