First let's draw this triangle:
Now, let's find the cosine of C using the law of cosines:
![\begin{gathered} c^2=a^2+b^2-2ab\cos (C) \\ 4^2=5^2+7^2-2\cdot5\cdot7\cdot\cos (C) \\ 16=25+49-70\cos (C) \\ 70\cos (C)=58 \\ \cos (C)=(58)/(70)=(29)/(35)=0.83 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/1ry191hxcnzwm63fdnuyht8sw1py38vmrx.png)
Now, in order to find the sin(A) and tan(C), let's first find the sin(C) using the property:
![\begin{gathered} \sin ^2(C)+\cos ^2(C)=1 \\ \sin ^2(C)+((29)/(35))^2=1 \\ \sin ^2(C)=1-(841)/(1225)=(384)/(1225) \\ \sin (C)=\text{0}.56 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/tgjaughfz2s2ljfcfbmhcyiwdkfc83j8do.png)
We can find the tan(C) using the relation:
![\begin{gathered} \tan (C)=(\sin (C))/(\cos (C)) \\ \tan (C)=(0.56)/(0.83)=0.67 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/mw8dzmn5i7qikqopngjo1bq5x6acu5g07u.png)
Finally, we can find sin(A) using the law of sines:
![\begin{gathered} (a)/(\sin(A))=(c)/(\sin (C)) \\ (5)/(\sin (A))=(4)/(0.56) \\ \sin (A)=0.56\cdot(5)/(4)=0.7 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/wa0aokarzxxsyotkcvallndgq34ld0gbax.png)