In the triangle shown, we have three angles: $\angle QRS$, $\angle RST$, and the exterior angle $\angle TSR$. Let me help break this down.

1. Finding the value of $x$:

The exterior angle $\angle TSR$ is equal to the sum of the two non-adjacent interior angles of the triangle (triangle angle sum property).

$\angle TSR = \angle QRS + \angle QSR$

We know:

• $\angle TSR = (6x + 18)^\circ$
• $\angle QRS = (x + 8)^\circ$
• $\angle QSR = (3x + 20)^\circ$

Substituting into the equation for the exterior angle:

$6x + 18 = (x + 8) + (3x + 20)$

Simplifying the right-hand side:

$6x + 18 = 4x + 28$

Now, solving for $x$:

$6x - 4x = 28 - 18$

$2x = 10$

$x = 5$

So, $x = 5$. You confirmed this earlier in your submission.

2. Solving for $\angle QRS$:

We substitute $x = 5$ into the expression for $\angle QRS$:

$\angle QRS = (x + 8)^\circ = (5 + 8)^\circ = 13^\circ$

Thus, $\angle QRS = 13^\circ$.

Would you like more details or have any further questions?

Related Questions:

1. What is the value of $\angle QSR$ using $x = 5$?
2. What is the sum of the angles inside the triangle $\triangle QRS$?
3. How does the exterior angle theorem help solve for unknown angles in a triangle?
4. Can you verify the interior angles of the triangle after substituting $x = 5$?
5. What are some other methods to find unknown angles in triangles?

Tip:

Always double-check whether a given angle is interior or exterior before applying theorems like the exterior angle theorem.