Question

in the figure above, $rt = tu$. what is the value of $x$?
a) 72
b) 66
c) 64
d) 58

Explanation:

Step1: Find angle at U in triangle RTU

Since \( RT = TU \), triangle \( RTU \) is isosceles with \( \angle R=\angle U \). The sum of angles in a triangle is \( 180^\circ \), and \( \angle RTU = 114^\circ \). So, \( \angle R+\angle U=180 - 114 = 66^\circ \). Since \( \angle R=\angle U \), \( \angle U=\frac{66}{2}=33^\circ \).

Step2: Find angle x using triangle SU R

In triangle \( SUR \), we know \( \angle S = 31^\circ \) and \( \angle U = 33^\circ \). The sum of angles in a triangle is \( 180^\circ \), but angle \( x \) is an exterior angle to triangle \( SUR \) at \( V \), so by exterior angle theorem, \( x=\angle S+\angle U \). Wait, no, actually, let's correct. Wait, the exterior angle at \( V \) (angle \( x \)) is equal to the sum of the two non - adjacent interior angles of triangle \( STU \)? Wait, no, let's re - examine.

Wait, first, in triangle \( RTU \), \( RT = TU \), so \( \angle TRU=\angle T UR \). \( \angle RTU = 114^\circ \), so \( \angle T UR=\frac{180 - 114}{2}=33^\circ \). Then, in triangle \( SUV \), the angle at \( T \) (vertical angle or adjacent? Wait, the angle at \( V \), \( x \), is an exterior angle for triangle \( S V U \)? Wait, no, let's use the exterior angle theorem properly.

The angle at \( T \) in triangle \( RTU \) is \( 114^\circ \), so the adjacent angle (linear pair) is \( 180 - 114 = 66^\circ \). Now, in triangle \( S V T \) (or the big triangle), we have \( \angle S = 31^\circ \), and the angle we just found ( \( 66^\circ \) )? Wait, no, let's start over.

Wait, the correct way: In triangle \( RTU \), \( RT = TU \), so it's isosceles. \( \angle RTU = 114^\circ \), so base angles \( \angle R=\angle U=\frac{180 - 114}{2}=33^\circ \). Then, in triangle \( S RU \), the sum of angles: \( \angle S=31^\circ \), \( \angle U = 33^\circ \), so the third angle (at \( V \)'s adjacent) is \( 180-(31 + 33)=116^\circ \), but that's not right. Wait, no, angle \( x \) is an exterior angle. Wait, the exterior angle at \( V \) (angle \( x \)) is equal to \( \angle S+\angle (angle at U in triangle RTU) \)? Wait, no, the exterior angle theorem states that an exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles.

Wait, let's look at the triangle formed by \( S \), \( V \), and \( U \). The angle at \( T \) (the angle supplementary to \( 114^\circ \)) is \( 66^\circ \). Then, in triangle \( S V U \), the angle at \( V \) (exterior angle \( x \)) is equal to \( \angle S+\angle (angle at U) \)? Wait, \( \angle S = 31^\circ \), and the angle at \( U \) in triangle \( RTU \) is \( 33^\circ \), no, wait, the angle at \( U \) in triangle \( SUR \) is \( 33^\circ \), and the angle at \( S \) is \( 31^\circ \), so the exterior angle at \( V \) (angle \( x \)) is \( 31+33 + 0\)? No, that's wrong. Wait, let's use the correct exterior angle.

Wait, the angle at \( T \) in triangle \( RTU \) is \( 114^\circ \), so the angle adjacent to it (linear pair) is \( 66^\circ \). Now, in triangle \( S V T \), we have angles: \( \angle S = 31^\circ \), and the angle we just found \( 66^\circ \)? No, that's not. Wait, I think I made a mistake earlier. Let's do it step by step.

1. In \( \triangle RTU \), \( RT = TU \), so \( \triangle RTU \) is isosceles with \( \angle R=\angle U \).
- \( \angle RTU = 114^\circ \)
- By angle - sum property of triangle, \( \angle R+\angle U+\angle RTU = 180^\circ \)
- \( 2\angle U+114 = 180 \)
- \( 2\angle U=180 - 114 = 66 \)
- \( \angle U = 33^\circ \)
2. Now, in \( \triangle SUV \), we want to find the exterior angle at \( V \) (which is \( x \)). The exterior angle theorem states that an exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles.
- The two non - adjacent interior angles to the exterior angle \( x \) are \( \angle S = 31^\circ \) and \( \angle U = 33^\circ \)? No, wait, no. Wait, the exterior angle at \( V \) (angle \( x \)) is equal to \( \angle S+\angle (angle at T in the other triangle) \)? Wait, no, let's look at the linear pair. The angle at \( T \) in \( \triangle RTU \) is \( 114^\circ \), so the angle supplementary to it (on the straight line \( SU \)) is \( 180 - 114 = 66^\circ \). Then, in the triangle formed by \( S \), \( V \), and the vertex with angle \( 66^\circ \), we have \( \angle S = 31^\circ \), and the angle \( 66^\circ \), so the exterior angle \( x \) is \( 31+66 - 180 \)? No, that's wrong.

Wait, I think the correct approach is:

The angle at \( T \) ( \( \angle RTU = 114^\circ \) ) and the angle adjacent to it (linear pair) is \( 66^\circ \). Now, in triangle \( S V T \), we have \( \angle S = 31^\circ \), and the angle \( 66^\circ \), so the angle at \( V \) (interior) is \( 180-(31 + 66)=83^\circ \), which is wrong.

Wait, let's start over.

Since \( RT = TU \), \( \triangle RTU \) is isosceles. \( \angle RTU = 114^\circ \), so \( \angle R=\angle U=\frac{180 - 114}{2}=33^\circ \).

Now, consider triangle \( S RU \). The sum of angles in a triangle is \( 180^\circ \). The angle at \( S \) is \( 31^\circ \), angle at \( U \) is \( 33^\circ \), so the angle at \( V \) (interior) is \( 180-(31 + 33)=116^\circ \), but that's not the exterior angle. Wait, no, angle \( x \) is an exterior angle, so \( x = 180 - 116=64^\circ \)? No, that's not matching the options. Wait, I must have messed up the diagram.

Wait, the diagram: \( S \) is the top vertex, \( R \) and \( U \) are at the base, \( T \) is on \( SU \), \( V \) is on \( RU \). \( RT = TU \), so triangle \( RTU \) with \( T \) on \( SU \), \( V \) on \( RU \).

So, \( \angle RTU = 114^\circ \), so the angle \( \angle STV = 180 - 114 = 66^\circ \) (linear pair). Now, in triangle \( STV \), we have \( \angle S = 31^\circ \), \( \angle STV = 66^\circ \), so the angle at \( V \) (interior) is \( 180-(31 + 66)=83^\circ \), no. Wait, the exterior angle at \( V \) (angle \( x \)) is equal to \( \angle S+\angle STV \)? Wait, \( \angle S = 31^\circ \), \( \angle STV = 66^\circ \), so \( x=31 + 66=97^\circ \), no.

Wait, I think the correct way is:

In triangle \( RTU \), \( RT = TU \), so \( \angle R=\angle U \). \( \angle RTU = 114^\circ \), so \( \angle U=(180 - 114)/2 = 33^\circ \).

Now, the angle at \( T \) ( \( \angle STU \)): since \( \angle RTU = 114^\circ \), \( \angle STU = 180 - 114 = 66^\circ \) (linear pair).

Now, in triangle \( SUV \), the exterior angle at \( V \) (angle \( x \)) is equal to \( \angle S+\angle STU \)? Wait, \( \angle S = 31^\circ \), \( \angle STU = 66^\circ \), so \( x = 31+66 - 180 \)? No, exterior angle theorem: exterior angle = sum of two non - adjacent interior angles.

Wait, the two non - adjacent interior angles to angle \( x \) are \( \angle S = 31^\circ \) and \( \angle U = 33^\circ \)? No, that gives \( 64^\circ \), which is option C. Wait, maybe I made a mistake in the first step.

Wait, let's recalculate \( \angle U \) in triangle \( RTU \):

\( \angle RTU = 114^\circ \), so \( \angle R+\angle U=180 - 114 = 66^\circ \). Since \( RT = TU \), \( \angle R=\angle U \), so \( \angle U = 33^\circ \). Then, in triangle \( S V U \), the exterior angle at \( V \) (angle \( x \)) is equal to \( \angle S+\angle U \)? No, \( \angle S = 31^\circ \), \( \angle U = 33^\circ \), sum is \( 64^\circ \), which is option C.

Wait, maybe the exterior angle theorem: the exterior angle at \( V \) is equal to \( \angle S+\angle (angle at T in triangle RTU's adjacent) \). Wait, no, let's use the fact that in triangle \( STU \), the sum of angles: \( \angle S = 31^\circ \), \( \angle U = 33^\circ \), so the angle at \( V \) (exterior) is \( 31+33 + 0\)? No, I think the correct answer is 64, which is option C. Wait, but let's check the steps again.

1. In \( \triangle RTU \), \( RT = TU \), so isosceles. \( \angle RTU = 114^\circ \).
- \( \angle R+\angle U=180 - 114 = 66^\circ \)
- \( \angle R=\angle U = 33^\circ \)
2. Now, consider the triangle formed by \( S \), \( V \), and \( U \). The angle at \( V \) (exterior angle \( x \)) is equal to \( \angle S+\angle U \)? No, \( \angle S = 31^\circ \), \( \angle U = 33^\circ \), sum is \( 64^\circ \). Yes, that's option C.

Answer:

C) 64