[tex]\qquad\qquad\huge\underline{{\sf Answer}} \huge♨[/tex]
In the given figure, NQ acts as a diameter and since diameter subtends 90° at the arc of the circle, so we can conclude that ~
[tex]\qquad \sf \dashrightarrow \:3y = 90 \degree[/tex]
[tex]\qquad \sf \dashrightarrow \:y = 90 \degree \div 3[/tex]
[tex]\qquad \sf \dashrightarrow \:y = 30 \degree [/tex]
Now, Let's use Angle sum property of Triangle to solve for z :
[tex]\qquad \sf \dashrightarrow \:4z + 3y + 50 \degree = 180 \degree[/tex]
[tex]\qquad \sf \dashrightarrow \:4z +90 \degree + 50 \degree = 180 \degree[/tex]
[tex]\qquad \sf \dashrightarrow \:4z = 180 \degree - (50 \degree + 90\degree)[/tex]
[tex]\qquad \sf \dashrightarrow \:4z = 180 \degree - (140\degree)[/tex]
[tex]\qquad \sf \dashrightarrow \:4z = 40 \degree[/tex]
[tex]\qquad \sf \dashrightarrow \:z = 40 \degree \div 4[/tex]
[tex]\qquad \sf \dashrightarrow \:z = 10 \degree [/tex]
I hope you understood the procedure, you can ask me in comments if you have any doubts.
