Answer:
26.4
Explanation:
Law Of Cosines:
![cos(A)=(b^2+c^2-a^2)/(2bc)](https://img.qammunity.org/2023/formulas/mathematics/high-school/20dxflj2zbv2y24a0iwdz7cn142e31x33r.png)
This should work for any side. This can generally be thought as:
![cos(\text{angle}) = \frac{\text{sum of squares of two other sides-opposite side squared}}{\text{2 times the product of the other two sides}}](https://img.qammunity.org/2023/formulas/mathematics/high-school/nhenfz64l6pv2riws24t368hat03ffulwj.png)
If this is too confusing here's the formula for the other sides (which is essentially the same, just different variables)
![cos(B)=(a^2+c^2-b^2)/(2ac)](https://img.qammunity.org/2023/formulas/mathematics/high-school/c2gky3ceaphuwn4ntgnse9iyxqt9k7v1z8.png)
![cos(C) =(a^2+b^2-c^2)/(2ab)](https://img.qammunity.org/2023/formulas/mathematics/high-school/d3fbqeb7av9r1xhh473me5zayu8jn3nuvx.png)
Anyways now just plug in the known values into the equation
![cos(A)=(4^2+6^2-3^2)/(2(6)(4))\\](https://img.qammunity.org/2023/formulas/mathematics/high-school/zs2ofqwp8lz3ycc23ufgfs06q1j0csnmih.png)
Square and multiply values
![cos(A)=(16+36-9)/(48)](https://img.qammunity.org/2023/formulas/mathematics/high-school/dgjh6rg3qftte5n4twgvnmvg6fq6hq59lx.png)
Add the values in the numerator
![cos(A)=(43)/(48)](https://img.qammunity.org/2023/formulas/mathematics/high-school/k8j3gg9ubs9bghiac2jcc6cq464wgw01pg.png)
Take the inverse of cosine on both sides
![A=cos^(-1)((43)/(48))](https://img.qammunity.org/2023/formulas/mathematics/high-school/z0kkpc0duik3a6kcxxpv0qbcathbsh7fyh.png)
calculate arccosine (inverse cosine) using a calculator
![A\approx 26.384](https://img.qammunity.org/2023/formulas/mathematics/high-school/3mof7ybrewn1om0mft08ydtbk8cxf3fi6a.png)
Round to nearest tenth
![A\approx26.4](https://img.qammunity.org/2023/formulas/mathematics/high-school/ws72m5qxw4gei8vn2d04h9iattyyvnuphi.png)