Let's make a drawing first
We have enough information to find the remaing angle in triangle QRS. The interior angles of a triangle add up 180º. So:
![180º=m\angle Q+84º+32º\Rightarrow m\angle Q=180º-84º-32º=64º](https://img.qammunity.org/2023/formulas/mathematics/college/7b83aa3bjock4cypwe0cy77d004i9smjnn.png)
Since both triangles are approximately the same, the measure of M is the same than the measure of Q:
![m\angle M\approx m\angle Q](https://img.qammunity.org/2023/formulas/mathematics/college/o92w1vd41acwbotbk9ijnz8wq86z7tzel2.png)
![17y-4=64º](https://img.qammunity.org/2023/formulas/mathematics/college/7qsl89rakqhejy8gmdtfe6g5jsvjesybjx.png)
Now we just clear y:
![y=(64+4)/(17)=4º](https://img.qammunity.org/2023/formulas/mathematics/college/5r9f9to3gzpmkitv9l77vyx646fq5uw4ve.png)
y = 4º
For the same reason, segments QS and MP are approximately the same too:
![15\approx2x+1](https://img.qammunity.org/2023/formulas/mathematics/college/h6nk0ylw70bzphb6r0fzevriuqjf67l0s2.png)
Then we clear x:
![x=(15-1)/(2)=7º](https://img.qammunity.org/2023/formulas/mathematics/college/ztgfrr0g0e16ij5w6yml8nuyxvcjzzutyc.png)
x = 7º