Question

Triangle Q R S is shown. Line R Q extends through point P. Angle Q S R is 35 degrees. Angle S R Q is 58 degrees. Exterior angle S Q P is x degrees.
What is the value of x?

[Art not to scale.]

x = 23
x = 35
x = 58
x = 93

Answer 1

To find the value of \( x \), which is the measure of the exterior angle \( S Q P \), we can use the properties of triangles and the exterior angle theorem.

In triangle \( QRS \):

1. We know:
- \( \angle QSR = 35^\circ \)
- \( \angle SRQ = 58^\circ \)

2. To find \( \angle QRS \), we can use the triangle sum property:
\[
\angle QRS + \angle QSR + \angle SRQ = 180^\circ
\]

3. Substituting the known angles:
\[
\angle QRS + 35^\circ + 58^\circ = 180^\circ
\]

4. Adding \( 35^\circ \) and \( 58^\circ \):
\[
35^\circ + 58^\circ = 93^\circ
\]

5. Now we have:
\[
\angle QRS + 93^\circ = 180^\circ
\]

6. Thus, we can find \( \angle QRS \):
\[
\angle QRS = 180^\circ - 93^\circ = 87^\circ
\]

Now, according to the exterior angle theorem, the exterior angle \( S Q P \) (which is \( x \)) is equal to the sum of the two opposite interior angles \( \angle QSR \) and \( \angle SRQ \):
\[
x = \angle QSR + \angle SRQ
\]
Substituting the values we found:
\[
x = 35^\circ + 58^\circ = 93^\circ
\]

Thus, the value of \( x \) is:
\[
\boxed{93}
\]