Question

Please see the picture below I NEED BOTH A AND B. (A&B) not just A

Answer 1

To solve this question, use the law of cosines shown below:

`c^2=a^2+b^2-2*a*b*cosC`

In this question,

a = 6

b = 8

C = 10°

First, let's find c:

`\begin{gathered} c^2=6^2+8^2-2*6*8*cos10 \\ c^2=36+64-94.54 \\ c^2=5.46 \\ √(c^2)=√(5.456) \\ c=2.34 \end{gathered}`

Second, let's find A.

Now, a, b, and c are known, and Cos A is not known:

`\begin{gathered} a^2=b^2+c^2-2bc*cosA \\ 6^2=8^2+2.34^2-2*8*2.34*cosA \\ 36=64+5.48-37.44*cosA \\ 36=69.48-37.44*cosA \\ \text{ Subtracting 69.46 from both sides:} \\ 36-69.48=69.46-37.44*cosA-69.46 \\ -33.48=-37.44*cosA \\ \text{ Dividing both sides by -37.44:} \\ (-33.48)/(-37.44)=(-37.44)/(-37.44)cosA \\ 0.8942=cosA \\ Then, \\ A=cos^(-1)(0.8942) \\ A=26.6 \end{gathered}`

Answer:

c = 2.34 units

A = 26.6°