Question

question 7 (1 point)
what is the value of c to the nearest tenth of a metre?
7 m
8 m
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 \(2\times7\times8\times\cos(50^{\circ})=2\times7\times8\times0.6428 = 89.312\).

$$c^{2}=49 + 64-89.312$$

$$c^{2}=113 - 89.312=23.688$$

Step3: Find \(c\)

Take the square - root of \(c^{2}\). \(c=\sqrt{23.688}\approx4.9\) (This is wrong. Let's correct the calculation).

Correct Step2:

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

$$c^{2}=49 + 64-2\times7\times8\times0.6428$$

$$c^{2}=49+64 - 70.4288$$

$$c^{2}=42.5712$$

Step3: Find \(c\)

$$c=\sqrt{42.5712}\approx6.525\approx6.4$$

(rounded to the nearest tenth)

Answer:

A. 6.4 m