Question

math 2201: acute triangle trigonometry september 2025 try it yourself try it yourself homework!! see google classroom!! if you need help, please reach out

Explanation:

Step1: Recall the Law of Cosines

The Law of Cosines formula for a triangle with sides \(a\), \(b\), \(c\) and the angle \(\theta\) opposite to side \(c\) is \(c^{2}=a^{2}+b^{2}-2ab\cos\theta\).

For the first triangle:

Let \(a = 20.4\), \(b = 23.6\), \(\theta=65^{\circ}\), and we want to find \(x\).

$$x^{2}=20.4^{2}+23.6^{2}-2\times20.4\times23.6\times\cos(65^{\circ})$$

$$x^{2}=416.16 + 556.96-963.84\times0.4226$$

$$x^{2}=416.16 + 556.96 - 407.44$$

$$x^{2}=565.68$$

$$x=\sqrt{565.68}\approx23.8$$

For the second triangle:

Let \(a = 73.9\), \(b = 89.5\), \(\theta = 45^{\circ}\), and we want to find \(x\).

$$x^{2}=73.9^{2}+89.5^{2}-2\times73.9\times89.5\times\cos(45^{\circ})$$

$$x^{2}=5461.21+8010.25 - 2\times73.9\times89.5\times\frac{\sqrt{2}}{2}$$

$$x^{2}=5461.21 + 8010.25-9317.91$$

$$x^{2}=4153.55$$

$$x=\sqrt{4153.55}\approx64.4$$

Answer:

For the first triangle, \(x\approx23.8\); for the second triangle, \(x\approx64.4\)