Question

jamal draws triangle xyz. if ( mangle x = 46^{circ}), ( xz = 10), and ( yz = 8), what is the approximate length of ( xy)?
3.4 or 10.4
6.4 or 13.7
10.4
13.7

Explanation:

Step1: Apply the Law of Cosines

The Law of Cosines formula is \(YZ^{2}=XZ^{2}+XY^{2}-2(XZ)(XY)\cos\angle X\). Let \(XY = x\). Substitute \(YZ = 8\), \(XZ=10\), and \(\angle X = 46^{\circ}\) (\(\cos46^{\circ}\approx0.6947\)) into the formula:

$$8^{2}=10^{2}+x^{2}-2\times10\times x\times0.6947$$

$$64 = 100+x^{2}-13.894x$$

$$x^{2}-13.894x + 36=0$$

Step2: Solve the quadratic equation

For a quadratic equation \(ax^{2}+bx + c = 0\) (\(a = 1\), \(b=- 13.894\), \(c = 36\)), use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\)

$$x=\frac{13.894\pm\sqrt{(-13.894)^{2}-4\times1\times36}}{2\times1}$$

$$x=\frac{13.894\pm\sqrt{192.04 - 144}}{2}$$

$$x=\frac{13.894\pm\sqrt{48.04}}{2}$$

$$x=\frac{13.894\pm6.93}{2}$$

Case1: Use the plus - sign

$$x=\frac{13.894 + 6.93}{2}=\frac{20.824}{2}=10.412\approx10.4$$

Case2: Use the minus - sign

$$x=\frac{13.894-6.93}{2}=\frac{6.964}{2}=3.482\approx3.4$$

Answer:

3.4 or 10.4