Question Show Examples In $\triangle X Y Z, x=1.2 \mathrm{~cm}, y=9.1 \mathrm{~cm}$ and $\angle Z=125^{\circ}$. Find the length of $z$, to the nearest 10 th of $a$ centimeter. Answer Attempt 1 out of 2 Submit Answer
Solution
Step 1 ：We are given a triangle XYZ with side lengths x = 1.2 cm, y = 9.1 cm, and an included angle Z = 125 degrees. We are asked to find the length of side z.
Step 2 ：We can use the Law of Cosines to solve for z. The Law of Cosines is given by the formula: \(c^2 = a^2 + b^2 - 2ab \cos(\gamma)\), where c is the length of the side opposite angle \(\gamma\), a and b are the lengths of the other two sides, and \(\gamma\) is the measure of the included angle.
Step 3 ：Substituting the given values into the formula, we get: \(z^2 = x^2 + y^2 - 2xy \cos(Z)\).
Step 4 ：Substituting the given values into the formula, we get: \(z^2 = (1.2)^2 + (9.1)^2 - 2(1.2)(9.1) \cos(125)\).
Step 5 ：Solving for z, we get: \(z = \sqrt{(1.2)^2 + (9.1)^2 - 2(1.2)(9.1) \cos(125)}\).
Step 6 ：Calculating the above expression, we get: \(z \approx 9.837525571499514\).
Step 7 ：Rounding to the nearest tenth of a centimeter, we get: \(z \approx 9.8\).
Step 8 ：Final Answer: The length of side \(z\) to the nearest tenth of a centimeter is \(\boxed{9.8}\) cm.
