Question

question 7 (1 point) what is the value of c to the nearest tenth of a metre? 50° 6.4 m 41.0 m 113.9 m 10.7 m

Explanation:

Step1: Apply the Law of Cosines

The Law of Cosines formula for a triangle with sides \(a\), \(b\), \(c\) and the included - angle \(C\) is \(c^{2}=a^{2}+b^{2}-2ab\cos C\). Here, \(a = 7\), \(b = 8\), and \(C = 50^{\circ}\).

$$c^{2}=7^{2}+8^{2}-2\times7\times8\times\cos(50^{\circ})$$

Step2: Calculate each term

First, \(7^{2}=49\), \(8^{2}=64\), and \(\cos(50^{\circ})\approx0.6428\). Then \(2\times7\times8\times\cos(50^{\circ})=2\times7\times8\times0.6428 = 89.22\).

$$c^{2}=49 + 64-89.22$$

$$c^{2}=113 - 89.22=23.78$$

Step3: Find the value of \(c\)

Take the square - root of \(c^{2}\). \(c=\sqrt{23.78}\approx4.9\) (This is wrong. Let's correct the calculation of \(2\times7\times8\times\cos(50^{\circ})\)).

$$2\times7\times8\times\cos(50^{\circ})=112\times0.6428 = 71.9936$$

$$c^{2}=49+64 - 71.9936=41.0064$$

$$c=\sqrt{41.0064}\approx6.4$$

Answer:

6.4 m