The diagram shows two parallel lines, $m$ and $n$, with transversals $s$ and $t$ intersecting them at point $P$. The angles formed at point $P$ include a 30° angle above line $m$ and a 70° angle below line $n$. The angle marked $x$ is also shown at point $P$.

Solution:

Given:

• $m \parallel n$
• $\angle Psm = 30^\circ$
• $\angle Ptn = 70^\circ$

Since $m$ and $n$ are parallel lines and $s$ and $t$ are transversals, the corresponding angles are equal.

Thus, $\angle Psm = \angle sP$ (corresponding angles), which gives $\angle sP = 30^\circ$.

Similarly, since $t$ is a transversal, $\angle tP$ is corresponding to $\angle Ptn = 70^\circ$.

We also know that the sum of angles around point $P$ on a straight line equals 180°.

$\angle sP + x + \angle tP = 180^\circ$

Substitute the known values:

$30^\circ + x + 70^\circ = 180^\circ$

$100^\circ + x = 180^\circ$

Solving for $x$:

$x = 180^\circ - 100^\circ$ $x = 80^\circ$

Final Answer:

The value of $x$ is $80^\circ$.

Would you like more details or have any further questions?

Follow-Up Questions:

1. What are corresponding angles, and why are they equal in this scenario?
2. Can you explain how the sum of angles on a straight line equals 180°?
3. How would the solution change if the given angles were different?
4. What is the relationship between alternate interior angles when lines are parallel?
5. How would you solve for $x$ if the lines $m$ and $n$ were not parallel?

Tip:

When dealing with parallel lines and a transversal, always remember that corresponding angles are equal, and the sum of angles on a straight line is 180°.