Question

Can you help me with number 15? Thank you I am having trouble with it.

Answer 1

x = 52.5° (option B)

Step-by-step explanation:

From the diagram: BC = 39, AB = 22. AC = 31

∠B = x

To get the value of x, we will apply cosine rule:

`\begin{gathered} b^2=a^2+c^2\text{ - 2acCosB} \\ \text{where b = AC, a = BC, c = AB} \\ \\ AC^2=BC^2+AB^2\text{ - 2(ABC)(AB)cosB} \end{gathered}`
`\begin{gathered} 31^2=39^2+22^2\text{ -2(39)(22)cosx} \\ 961\text{ = 1521 + 484 -1716cosx} \\ 961\text{ = 2005 -1716cosx} \end{gathered}`
`\begin{gathered} 961\text{ = 2005 -1716cosx} \\ 961\text{ - 2005 = -1716cosx} \\ -1044\text{ = -1716cosx} \\ \\ \text{divide both sides by -1716:} \\ \frac{-1044\text{ }}{\text{-1716}}\text{= }\frac{\text{-1716cosx}}{\text{-1716}} \\ \text{cos x = 0.6084} \\ x=cos^(-1)(\text{0.6084)} \\ x\text{ = 52.53}\degree \end{gathered}`

To the nearest tenth, x = 52.5° (option B)