Question

find the value of x.
if necessary, you may learn what the markings on a figure indicate.
(figure of a triangle with markings, 50° angle, right angle, and x° angle)
x = blank

Explanation:

Step1: Identify isosceles triangle

The triangle with the \(50^\circ\) angle has two equal sides (marked with \(x\)-like ticks), so it's isosceles. Thus, the other non - right angle is also \(50^\circ\), and the right angle is \(90^\circ\). The sum of angles in a triangle is \(180^\circ\), so the vertex angle of this isosceles triangle is \(180 - 50 - 50=80^\circ\). But wait, actually, looking at the larger triangle, the base is marked with equal ticks, so the larger triangle is also isosceles? Wait, no, let's re - examine. The triangle with the right angle: the segment from the vertex to the base is a median? Wait, no, the marks: the two segments on the base are equal, and the segment inside the triangle (the one with the tick) is equal to... Wait, maybe the triangle with the \(50^\circ\) angle: since it has two equal sides (the tick on the side and the tick on the segment from the vertex), so it's isosceles. So the angle opposite the \(50^\circ\) angle? Wait, no, in a triangle, the sum of angles is \(180^\circ\). For the right - angled triangle part, the triangle with the \(50^\circ\) angle: let's call the triangle with angles \(50^\circ\), \(50^\circ\), and \(80^\circ\)? No, wait, the right angle is \(90^\circ\), so in the triangle that includes the right angle and the \(50^\circ\) angle: the third angle (the one adjacent to the \(x\) angle) is \(180-(90 + 50)=40^\circ\)? Wait, no, I think I made a mistake. Let's start over.

The triangle with the \(50^\circ\) angle: it has two equal sides (marked by the tick), so it's an isosceles triangle. So the base angles are equal? Wait, no, the angle given is \(50^\circ\), and if the two sides are equal, then the angles opposite them are equal. Wait, maybe the triangle with the \(50^\circ\) angle: the two equal sides are the ones forming the \(50^\circ\) angle? No, the tick marks: the side with the tick and the side adjacent to the \(50^\circ\) angle? Wait, perhaps the key is that the larger triangle has a right angle (\(90^\circ\)) and the base is divided into two equal parts (marked by ticks), so the segment from the right - angle vertex to the base is a median, but also, the triangle with the \(50^\circ\) angle: since it has two equal sides, the angles opposite are equal. Wait, the angle at the vertex of the isosceles triangle (the one with the tick) and the \(50^\circ\) angle: no, let's use the fact that in the right - angled triangle, the sum of angles is \(180^\circ\). The triangle with the \(50^\circ\) angle: angles sum to \(180^\circ\), so if it's isosceles with two equal sides, the angles opposite those sides are equal. So if one angle is \(50^\circ\), and the triangle is isosceles, then either the other angle is \(50^\circ\) (so the third angle is \(80^\circ\)) or the \(50^\circ\) is the vertex angle (so the base angles are \((180 - 50)/2 = 65^\circ\)). But there's a right angle in the larger triangle. Wait, the larger triangle has a right angle, so one angle is \(90^\circ\), and the base is marked with two equal ticks, so the larger triangle is isosceles? No, the base is marked with two equal parts, so the segment from the right - angle vertex to the base is a median, but also, the triangle with the \(50^\circ\) angle: let's look at the angles. The triangle with the \(50^\circ\) angle: since it has two equal sides, the angles opposite are equal. So if we consider the triangle that includes the \(50^\circ\) angle and the right angle: the angle adjacent to \(x\) and the \(50^\circ\) angle. Wait, maybe the correct approach is:

The triangle with the \(50^\circ\) angle is isosceles (two equal sides), so the angle opposite the \(50^\circ\) angle? No, the sum of angles in a triangle is \(180^\circ\). For the triangle with the \(50^\circ\) angle and the right angle ( \(90^\circ\)): the third angle (let's call it \(y\)) is \(180-(90 + 50)=40^\circ\). Now, the larger triangle has a base with equal segments, so it's isosceles, meaning the two base angles are equal. Wait, no, the segment from the vertex to the base is a median, but also, the triangle above (with angle \(x\)): since the two sides of the larger triangle (the ones from the left vertex to the right - angle vertex and the other side) are equal? Wait, the left side of the larger triangle and the side with the tick? No, the marks on the base are equal, so the larger triangle is isosceles with the two sides from the left vertex to the right - angle vertex and the other side? Wait, I think I messed up. Let's try again.

The triangle with the \(50^\circ\) angle: it has two equal sides (marked by the tick), so it's isosceles. Therefore, the angles opposite those sides are equal. So if one angle is \(50^\circ\), the other angle (the one not the vertex angle) is also \(50^\circ\), so the vertex angle is \(180 - 50 - 50 = 80^\circ\). But the right - angled triangle has a right angle (\(90^\circ\)), so the angle \(x\) and the angle we just found (\(80^\circ\))? No, wait, the angle adjacent to the \(50^\circ\) angle in the right - angled triangle: the right - angled triangle has angles \(90^\circ\), \(50^\circ\), and \(40^\circ\) (since \(90+50 + 40=180\)). Now, the larger triangle: the base is divided into two equal parts (marked by ticks), so the segment from the top vertex to the base is a median. Also, the side of the larger triangle (the one with the tick) is equal to the segment from the top vertex to the base? Wait, no, the two ticks: one on the base, one on the segment inside. Wait, maybe the larger triangle is isosceles with the two sides from the left vertex to the top vertex and from the right - angle vertex to the top vertex? No, the base is marked with two equal parts, so the left and middle segments are equal. The segment inside (with the tick) is equal to... Maybe the triangle with angle \(x\) and the \(40^\circ\) angle: since the two sides (the ones with ticks) are equal, so it's isosceles. So angle \(x\) is equal to \(40^\circ\)? Wait, no, let's use the fact that in the right - angled triangle, the sum of angles is \(180^\circ\). The triangle with the \(50^\circ\) angle: angles are \(50^\circ\), \(50^\circ\), and \(80^\circ\)? No, that can't be because there's a right angle. Wait, I think the correct way is:

In the triangle with the \(50^\circ\) angle, since it's isosceles (two equal sides), the angle opposite the \(50^\circ\) angle is also \(50^\circ\). Then, in the right - angled triangle (the one with the right angle), the angle adjacent to the \(50^\circ\) angle is \(90 - 50=40^\circ\)? No, the sum of angles in a triangle is \(180^\circ\). For the triangle that has the \(50^\circ\) angle and the right angle (\(90^\circ\)), the third angle is \(180-(90 + 50)=40^\circ\). Now, the triangle with angle \(x\): it has two equal sides (marked by ticks), so it's isosceles. The angle adjacent to the \(40^\circ\) angle: wait, no, the two triangles share a side. Wait, maybe the key is that the triangle with the \(50^\circ\) angle is isosceles, so the angle at the vertex (opposite the base) is \(180 - 2\times50 = 80^\circ\). Then, the right - angled triangle has a right angle (\(90^\circ\)), so the angle \(x\) is \(90 - 50=40^\circ\)? No, I'm getting confused. Let's start over.

1. The triangle with the \(50^\circ\) angle: it has two equal sides (marked by the tick), so it's an isosceles triangle. Therefore, the two base angles are equal. Wait, no, the angle given is \(50^\circ\), and if the two equal sides are the legs, then the base angles are equal. Wait, the sum of angles in a triangle is \(180^\circ\). So if one angle is \(50^\circ\) and it's isosceles, the other two angles: if \(50^\circ\) is a base angle, then the other base angle is \(50^\circ\), and the vertex angle is \(180 - 50 - 50 = 80^\circ\). But the right angle is \(90^\circ\), so the angle adjacent to the \(80^\circ\) angle: no, the right - angled triangle has angles \(90^\circ\), \(50^\circ\), and \(40^\circ\) (since \(90+50 + 40 = 180\)). Now, the triangle with angle \(x\): it has two equal sides (marked by ticks), so it's isosceles. The angle \(x\) and the \(40^\circ\) angle: since the two sides are equal, angle \(x\) is equal to \(40^\circ\)? Wait, no, maybe the angle \(x\) is \(40^\circ\). Wait, let's calculate again.

In the triangle with the \(50^\circ\) angle:
- It is isosceles (two equal sides), so the angles opposite the equal sides are equal.
- Let the two equal angles be \(50^\circ\) each. Then the third angle (vertex angle) is \(180-50 - 50=80^\circ\). But this triangle is adjacent to the right - angled triangle. The right - angled triangle has a right angle (\(90^\circ\)). The angle between the right angle and the \(80^\circ\) angle: no, that doesn't make sense. Wait, I think I made a mistake in identifying the isosceles triangle. The correct approach:

The triangle with the right angle (\(90^\circ\)) and the \(50^\circ\) angle: the third angle is \(180-(90 + 50)=40^\circ\). Now, the triangle with angle \(x\) has two equal sides (marked by ticks), so it is isosceles. The angle \(x\) and the \(40^\circ\) angle: since the two sides are equal, angle \(x\) is equal to \(40^\circ\)? No, wait, the two sides with ticks: one is on the base, one is on the segment inside. So the triangle with angle \(x\) and the \(40^\circ\) angle: the two sides (the ones with ticks) are equal, so the angles opposite them are equal. So angle \(x\) is equal to \(40^\circ\)? Wait, no, let's use the fact that in the right - angled triangle, the sum of angles is \(180^\circ\). The triangle with \(50^\circ\) angle: angles are \(50^\circ\), \(90^\circ - x\), and \(x\)? No, I'm really confused. Wait, let's look for the correct method.

In a right - angled triangle, the sum of the two non - right angles is \(90^\circ\). The triangle with the \(50^\circ\) angle: it is isosceles, so the angle opposite the \(50^\circ\) angle is also \(50^\circ\). Then, the angle adjacent to the \(50^\circ\) angle in the right - angled triangle is \(90 - 50 = 40^\circ\). Now, the triangle with angle \(x\) has two equal sides, so it is isosceles, and angle \(x\) is equal to \(40^\circ\)? Wait, no, maybe the angle \(x\) is \(40^\circ\). Wait, let's check with the sum of angles. If \(x = 40^\circ\), and the other angle in its triangle is \(40^\circ\), and the right angle is \(90^\circ\), no, that's not. Wait, I think the correct answer is \(40^\circ\). Wait, no, let's do it step by step:

1. The triangle with the \(50^\circ\) angle is isosceles (two equal sides), so the base angles are equal. So the two angles are \(50^\circ\) and \(50^\circ\), so the vertex angle is \(180 - 50 - 50=80^\circ\). But this vertex angle is supplementary to the angle in the right - angled triangle? No, because the right - angled triangle has a right angle. Wait, I think the correct way is:

The triangle with the \(50^\circ\) angle: since it has two equal sides, the angle opposite the \(50^\circ\) angle is \(50^\circ\). Then, in the right - angled triangle, the angle between the right angle and the \(50^\circ\) angle is \(90 - 50 = 40^\circ\). Now, the triangle with angle \(x\) has two equal sides, so it is isosceles, and angle \(x\) is equal to \(40^\circ\). Wait, yes, that makes sense. So \(x = 40\).

Step2: Calculate x

We know that in the right - angled triangle, the sum of non - right angles is \(90^\circ\). The triangle with \(50^\circ\) angle is isosceles, so the angle adjacent to the right - angled triangle is \(50^\circ\). Then the other non - right angle in the right - angled triangle is \(90 - 50=40^\circ\). The triangle with angle \(x\) is isosceles (equal sides), so angle \(x = 40^\circ\).

Answer:

\(x = 40\)