In the given image, we have a triangle $\triangle QRS$ with angles labeled as expressions involving $x$. The exterior angle at $S$, labeled $3x - 28$, corresponds to the exterior angle theorem. The two interior opposite angles are $x + 7$ and $x + 15$.

According to the exterior angle theorem, the exterior angle is equal to the sum of the two non-adjacent interior angles. Therefore, we can set up the equation:

$3x - 28 = (x + 7) + (x + 15)$

Now, let’s solve for $x$:

1. Combine like terms on the right-hand side: $3x - 28 = 2x + 22$

2. Subtract $2x$ from both sides: $x - 28 = 22$

3. Add 28 to both sides: $x = 50$

Thus, $x = 50$.

To verify:

• The interior angles are $x + 7 = 50 + 7 = 57^\circ$ and $x + 15 = 50 + 15 = 65^\circ$.
• The exterior angle is $3x - 28 = 3(50) - 28 = 150 - 28 = 122^\circ$.

Check: $57^\circ + 65^\circ = 122^\circ$, which matches the exterior angle, confirming the solution is correct.

Would you like more details on any part, or do you have any other questions?

Related questions:

1. What is the exterior angle theorem and how is it applied in different cases?
2. How would the solution change if one of the angles was altered?
3. Can this problem be extended to a polygon with more sides?
4. What are the conditions for solving exterior angle problems in non-triangular polygons?
5. How can you calculate the sum of all interior angles of a triangle?

Tip:

Always use the exterior angle theorem when dealing with problems involving a triangle and its extended sides. This simplifies finding unknown angles.