Question

plane geometry 14 (luke barsky, id: 14176440)
in the figure, solve for x and y.
draw your picture, identify all triangles within your picture
diagram with angles 68°, 85°, 74°, 50°, and angles x° and y° on a straight line
x =
y =

Explanation:

Step1: Find the third angle of the triangle with 50° and 85°

The sum of angles in a triangle is \(180^\circ\). So the third angle \(z = 180 - 50 - 85 = 45^\circ\).

Step2: Solve for \(x\)

In the triangle with \(68^\circ\) and \(45^\circ\), the sum of angles is \(180^\circ\). So \(x = 180 - 68 - 45 = 67^\circ\)? Wait, no, wait. Wait, the straight line is \(180^\circ\). Wait, maybe I made a mistake. Wait, the triangle with \(68^\circ\) and \(x\): Wait, the vertical angle to the 45° angle (from the triangle with 50 and 85) is also 45°, because vertical angles are equal. So in the left triangle, angles are 68°, 45°, and \(x\)? No, wait, \(x\) is an exterior angle? Wait, no, the left triangle: the angles inside are 68°, and the angle adjacent to \(x\) is part of the straight line. Wait, let's re-examine.

Wait, the triangle with 50° and 85°: the third angle is \(180 - 50 - 85 = 45^\circ\). Then, the vertical angle to this 45° is also 45° (vertical angles are equal). Now, in the left triangle, we have angles 68°, 45°, and the angle adjacent to \(x\). Wait, no, \(x\) is an exterior angle? Wait, no, the left triangle: the angles inside are 68°, and the angle opposite to \(x\)? Wait, no, the straight line is \(180^\circ\). So \(x = 180 - 68 - 45\)? Wait, no, that would be the interior angle. Wait, no, \(x\) is an exterior angle? Wait, no, the left triangle: the angles are 68°, 45°, and the angle at the base (adjacent to \(x\)). Then, \(x = 180 - (68 + 45) = 67^\circ\)? Wait, no, that's the interior angle. Wait, no, \(x\) is on the straight line, so the interior angle of the triangle is \(180 - x\). Wait, I think I messed up. Let's start over.

First, the triangle with angles 50° and 85°: third angle is \(180 - 50 - 85 = 45^\circ\). This angle is vertical to the angle in the left triangle (the triangle with 68°). So the left triangle has angles 68°, 45°, and the angle at the base (let's call it \(a\)). So \(a = 180 - 68 - 45 = 67^\circ\). Then, since \(a\) and \(x\) are supplementary (they form a straight line), \(x = 180 - 67 = 113^\circ\)? Wait, that can't be. Wait, no, maybe \(x\) is an interior angle. Wait, the problem says "solve for \(x\) and \(y\)". Let's look at the right triangle: it has 74°, and the angle adjacent to \(y\).

Wait, the triangle with 50° and 85°: third angle is 45°, vertical angle is 45°, so in the right triangle, the angles are 74°, 45°, and the angle adjacent to \(y\). Then, that angle is \(180 - 74 - 45 = 61^\circ\), so \(y = 180 - 61 = 119^\circ\)? No, this is confusing. Wait, maybe the left triangle: angles are 68°, and the angle opposite to \(x\) is 45°? Wait, no, let's use the exterior angle theorem.

Wait, the triangle with 50° and 85°: the third angle is 45°, vertical angle is 45°. Then, in the left triangle, the exterior angle \(x\) is equal to the sum of the two non-adjacent interior angles? Wait, no, the left triangle has angles 68° and 45°, so the exterior angle \(x\) would be \(180 - (68 + 45) = 67^\circ\)? No, that's not right. Wait, maybe I made a mistake in the vertical angle. Wait, the triangle with 50° and 85°: the third angle is \(180 - 50 - 85 = 45^\circ\). Then, the angle adjacent to the left triangle is 45° (vertical angle). So in the left triangle, the angles are 68°, 45°, and the angle at the base (let's call it \(b\)). So \(b = 180 - 68 - 45 = 67^\circ\). Then, since \(b\) and \(x\) are supplementary (they form a straight line), \(x = 180 - 67 = 113^\circ\)? Wait, that seems high. Wait, maybe the left triangle is: angle at the top is 68°, the angle at the bottom (adjacent to \(x\)) is \(x\), and the angle in the middle is 45° (vertical angle). Wait, no, the sum of angles in a triangle is 180, so \(68 + 45 + x = 180\)? No, \(x\) is not inside the triangle. Wait, \(x\) is on the straight line, so the interior angle of the triangle is \(180 - x\). So \(68 + 45 + (180 - x) = 180\)? No, that can't be. I think I need to re-express.

Wait, let's identify the triangles:

1. Triangle 1: angles 50°, 85°, and 45° (since 50 + 85 + 45 = 180).

2. Triangle 2: angles 68°, 45°, and (let's call it \(c\)). So \(68 + 45 + c = 180\) → \(c = 180 - 68 - 45 = 67\). Then, \(x\) is supplementary to \(c\), so \(x = 180 - 67 = 113\)? Wait, no, that would mean \(x\) is 113°, but let's check the right triangle.

Triangle 3: angles 74°, 45°, and (let's call it \(d\)). So \(74 + 45 + d = 180\) → \(d = 180 - 74 - 45 = 61\). Then, \(y\) is supplementary to \(d\), so \(y = 180 - 61 = 119\)? Wait, but that seems odd. Wait, maybe the 45° angle is not the vertical angle. Wait, the triangle with 50° and 85°: the angle adjacent to the right triangle is 50°? No, the 50° angle is at the base. Wait, maybe the vertical angle to the 85° angle? No, 85° is one angle, 50° is another, so the third is 45°, which is vertical to the angle in the left and right triangles. Wait, maybe the left triangle has angles 68°, and the angle equal to 45° (vertical angle), so the third angle is \(180 - 68 - 45 = 67\), and \(x\) is equal to that third angle? No, \(x\) is on the straight line. Wait, I'm confused. Let's try again.

Wait, the key is that the sum of angles in a triangle is 180°, and vertical angles are equal.

First, the triangle with angles 50° and 85°: the third angle is \(180 - 50 - 85 = 45^\circ\). This angle is vertical to the angle in the left triangle (so 45°) and also vertical to the angle in the right triangle (so 45°).

Now, in the left triangle (with 68°), the angles are 68°, 45°, and the angle adjacent to \(x\). Wait, no, \(x\) is an exterior angle. Wait, the exterior angle theorem: the exterior angle is equal to the sum of the two non-adjacent interior angles. So in the left triangle, the exterior angle \(x\) would be equal to 68° + 45° = 113°? Wait, that's the exterior angle theorem. Yes! Because the exterior angle \(x\) is equal to the sum of the two remote interior angles (68° and 45°). So \(x = 68 + 45 = 113^\circ\).

Then, in the right triangle, the exterior angle \(y\) is equal to the sum of the two remote interior angles: 74° and 45°. So \(y = 74 + 45 = 119^\circ\). Wait, but let's check with the triangle sum.

In the left triangle, the interior angle adjacent to \(x\) is \(180 - x = 180 - 113 = 67^\circ\). Then, 68 + 45 + 67 = 180? 68 + 45 is 113, 113 + 67 is 180. Yes, that works.

In the right triangle, the interior angle adjacent to \(y\) is \(180 - y = 180 - 119 = 61^\circ\). Then, 74 + 45 + 61 = 180? 74 + 45 is 119, 119 + 61 is 180. Yes, that works.

So:

Step1: Find the third angle of the triangle with 50° and 85°

The third angle is \(180 - 50 - 85 = 45^\circ\) (let's call this angle \(A\)).

Step2: Solve for \(x\) using the exterior angle theorem

In the left triangle, the exterior angle \(x\) is equal to the sum of the two non-adjacent interior angles: \(68^\circ\) and \(45^\circ\). So \(x = 68 + 45 = 113^\circ\).

Step3: Solve for \(y\) using the exterior angle theorem

In the right triangle, the exterior angle \(y\) is equal to the sum of the two non-adjacent interior angles: \(74^\circ\) and \(45^\circ\). So \(y = 74 + 45 = 119^\circ\).

Answer:

\(x = 113^\circ\), \(y = 119^\circ\)