Let's solve point a of the question. To calculate the diameter of the figure we can solve it using the Pythagorean theorem
![\begin{gathered} NK=√(NL^2+KL^2) \\ NK=√(12^2+5^2) \\ NK=√(144+25) \\ NK=√(169) \\ NK=13 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/u8jpj63il4a369jrqam2cp1op1qnj4ps8n.png)
Let us now calculate part b.
![mNLK=180](https://img.qammunity.org/2023/formulas/mathematics/college/wl4pvayu70q512lz6ogfp2wlbkmy6hcwqb.png)
Let us now calculate part c.
![mNJK=180](https://img.qammunity.org/2023/formulas/mathematics/college/eltcydocnjjqie34vttd1wqho0j7i0k5g1.png)
Now let us calculate the arc for the part d. As it is a right triangle it would be:
![m\angle NLK=90](https://img.qammunity.org/2023/formulas/mathematics/college/uv8y9ugnmpum114us8jt98jlkvp9nqs4d9.png)
Now let us calculate for part e
![mKL=45.2](https://img.qammunity.org/2023/formulas/mathematics/college/klgasyonrzdn80ib0ubj6wxesezeita6du.png)
Finally, for part f
![\begin{gathered} mNL=180-45.2 \\ mNL=134.8 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/70cd6ybx0svrbsd6ocij0cej9vc2ui4zjc.png)