Question

find the measures of the angles marked x and y. remember that (1) the sum of the measures of the angles of a triangle is 180° and (2) supplementary angles have a sum of 180°. x = \square and y = \square (simplify your answer. do not include the degree symbol in your answer.)

Explanation:

Step1: Analyze the right triangle

The triangle is a right triangle, so one angle is \(90^\circ\), another is \(x^\circ\), and the third angle (interior angle adjacent to \(y\)) and \(y\) are supplementary. Also, the exterior angle \((2x + 50)^\circ\) and the interior angle adjacent to it are supplementary. First, use the fact that in a right triangle, the two non - right angles are related to the exterior angle. The exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles. So, \(2x + 50=90 + x\).

Step2: Solve for x

Solve the equation \(2x+50 = 90+x\). Subtract \(x\) from both sides: \(2x - x+50=90+x - x\), which gives \(x + 50=90\). Then subtract 50 from both sides: \(x=90 - 50=40\).

Step3: Solve for y

Now that we know \(x = 40\), the angle \(2x + 50=2\times40+50 = 80 + 50=130\). Since \(y\) and \(2x + 50\) are supplementary (they form a linear pair), \(y+2x + 50=180\). Substitute \(2x + 50 = 130\) into the equation: \(y+130 = 180\). Subtract 130 from both sides: \(y=180 - 130 = 50\). Or we can use the triangle angle sum. The interior angle adjacent to \(y\) is \(180 - y\). In the right triangle, \(90+x+(180 - y)=180\). Substitute \(x = 40\): \(90 + 40+180 - y=180\), \(310 - y=180\), \(y=310 - 180 = 130\)? Wait, no, I made a mistake. Wait, the exterior angle is equal to the sum of the two non - adjacent interior angles. The two non - adjacent interior angles of the exterior angle \((2x + 50)^\circ\) are the right angle (\(90^\circ\)) and \(x^\circ\). So \(2x+50=90 + x\) (correct). Then \(x = 40\). Now, the angle \(y\) and the interior angle at the base of the triangle (let's call it \(z\)) are supplementary, and \(z=180-(90 + x)=180-(90 + 40)=50\). So \(y = 180 - z=180 - 50 = 130\)? Wait, no, I messed up the exterior angle. Wait, the exterior angle is \(2x + 50\), and the interior angle adjacent to it is \(180-(2x + 50)\). In the triangle, the sum of angles is \(90+x+(180-(2x + 50))=180\). Let's expand: \(90+x + 180-2x - 50=180\), \(220 - x=180\), \(x=220 - 180 = 40\) (same as before). Then the interior angle adjacent to \(y\) is \(180-(2x + 50)=180-(80 + 50)=50\). Then, since \(y\) and this interior angle are supplementary, \(y = 180 - 50=130\)? Wait, now I'm confused. Wait, let's draw the triangle. It's a right - angled triangle, right angle at the bottom left. The base is horizontal, the other side is vertical. The hypotenuse is from the top left (right angle) to the bottom right. Then, there is an extension of the base to the right, forming an exterior angle at the bottom right vertex. So the interior angle at the bottom right of the triangle is \(180 - y\) (since \(y\) is the exterior angle there). In the triangle, the sum of angles is \(90 + x+(180 - y)=180\). So \(90+x + 180-y=180\), \(x - y=- 90\), \(y=x + 90\). But we also know that the exterior angle \(2x + 50\) and \(y\) are supplementary? No, the exterior angle and the interior angle at that vertex are supplementary. So \(2x + 50+(180 - y)=180\), \(2x+50 - y=0\), \(y = 2x+50\). But from the triangle angle sum: \(90+x+(180 - y)=180\), \(x - y=- 90\), \(y=x + 90\). So set \(2x + 50=x + 90\), \(2x - x=90 - 50\), \(x = 40\). Then \(y=x + 90=40+90 = 130\). Ah, I see my earlier mistake. The exterior angle is equal to the sum of the two non - adjacent interior angles. The two non - adjacent interior angles to the exterior angle \(2x + 50\) are \(x\) and \(90\), so \(2x+50=x + 90\) (correct). Then \(x = 40\), and \(y\) is supplementary to the interior angle at that vertex. The interior angle at that vertex is \(180 - y\), and in the triangle, \(90+x+(180 - y)=180\), so \(x - y=- 90\), \(y=x + 90=130\). Or, since \(y\) and \(2x + 50\) are supplementary? No, \(y\) and the interior angle (let's call it \(z\)) are supplementary, and \(z\) and \(2x + 50\) are supplementary? No, \(z\) and \(2x + 50\) are a linear pair, so \(z+2x + 50=180\), and \(z = 180-(2x + 50)\). In the triangle, \(90+x+z=180\), so \(90+x+180-(2x + 50)=180\), \(90+x + 180-2x - 50=180\), \(220 - x=180\), \(x = 40\). Then \(z=180-(2\times40 + 50)=180 - 130 = 50\). Then \(y=180 - z=180 - 50 = 130\). Yes, that's correct. My earlier mistake was in the relationship between \(y\) and the interior angle. So \(x = 40\), \(y = 130\). Wait, but let's check with the exterior angle formula. The exterior angle is equal to the sum of the two non - adjacent interior angles. The two non - adjacent interior angles are \(x\) (top angle) and \(90\) (right angle). So exterior angle \(=x + 90\). But the exterior angle is given as \(2x + 50\). So \(2x+50=x + 90\), so \(x = 40\). Then exterior angle \(=2\times40+50 = 130\). But \(y\) and the exterior angle are supplementary? No, \(y\) is the angle adjacent to the exterior angle? Wait, no, looking at the diagram: the base is horizontal, the right angle is at the bottom left. The hypotenuse goes from bottom left to top right? Wait, no, the right angle is at the bottom left, so the vertical side is on the left, horizontal base on the bottom, and hypotenuse from top left (right angle) to bottom right. Then, the base is extended to the right, so the angle at the bottom right between the hypotenuse and the extended base is \(y\), and the angle between the extended base and the hypotenuse's extension is \(2x + 50\). So \(y\) and \(2x + 50\) are supplementary? No, \(y\) and the interior angle (between hypotenuse and base) are supplementary? Wait, I think I mis - identified the angles. Let's start over.

In a right triangle, one angle is \(90^\circ\), another is \(x^\circ\), and the third angle (let's call it \(z^\circ\)) is such that \(90 + x+z = 180\), so \(x + z=90\), so \(z = 90 - x\). Now, the angle \(z\) and \(y\) are supplementary (because they form a linear pair with the angle \(2x + 50\)? Wait, no, \(z\), \(y\), and \(2x + 50\): \(z + y+2x + 50=360\)? No, that's not right. Wait, the three angles on a straight line sum to \(180^\circ\). So \(z + y+(2x + 50)=180\)? No, no, the angle \(z\) (interior angle at bottom right of triangle), \(y\), and \(2x + 50\) are on a straight line? No, \(z\) is the interior angle at bottom right of the triangle, \(y\) is the angle between the hypotenuse and the extended base, and \(2x + 50\) is the angle between the extended base and the other side. So \(y\) and \(2x + 50\) are supplementary? No, \(y+2x + 50 = 180\) (since they form a linear pair). And \(z + y=180\) (since \(z\) and \(y\) form a linear pair). So \(z=2x + 50\). But we also know that in the right triangle, \(90 + x+z = 180\), so \(x+z = 90\). Substitute \(z = 2x + 50\) into \(x+z = 90\): \(x+2x + 50=90\), \(3x+50 = 90\), \(3x=40\), \(x=\frac{40}{3}\)? No, that's wrong. Wait, I think the correct approach is: 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 exterior angle here is \(2x + 50\), and the two non - adjacent interior angles are the right angle (\(90^\circ\)) and \(x^\circ\). So \(2x + 50=90 + x\). Solving for \(x\): \(2x - x=90 - 50\), \(x = 40\). Now, the angle \(y\) and the exterior angle \(2x + 50\) are supplementary? No, \(y\) is the angle adjacent to the exterior angle? Wait, no, \(y\) is the angle between the hypotenuse and the extended base, and the exterior angle is between the extended base and the other side. So \(y\) and the interior angle (between hypotenuse and base) are supplementary, and the interior angle (between hypotenuse and base) is equal to \(180-(2x + 50)\) (since they are supplementary). In the triangle, \(90 + x+(180-(2x + 50))=180\). Simplify: \(90+x + 180-2x - 50=180\), \(220 - x=180\), \(x = 40\). Then the interior angle (between hypotenuse and base) is \(180-(2\times40 + 50)=180 - 130 = 50\). Then \(y = 180 - 50=130\) (since \(y\) and this interior angle are supplementary). Yes, that's correct. So \(x = 40\), \(y = 130\).

Answer:

\(x = 40\) and \(y = 130\)