Triangles, Lines, & Angles - SAT Mathematics - Free Printable

Let’s solve this step by step.

We are given a diagram with several angles labeled, and we need to find the value of x°.

First, let’s look at the triangle that has angles: 20°, 40°, and an unknown angle at the bottom left (let’s call it A).

In any triangle, the sum of interior angles is always 180°.

So for that top triangle:
20° + 40° + A = 180°
→ A = 180° - 60° = 120°

But wait — in the diagram, there’s already an angle labeled 120° near the center. Let’s check if that matches.

Actually, looking again: the 120° angle is inside the figure where two lines cross — it might be part of a different triangle or a vertical angle situation.

Let me try another approach.

Look at the big triangle on the right side — it has angles: 15°, and part of x°, and maybe the 120°? Not quite.

Wait — perhaps we can use the fact that angles around a point add up to 360°, or that straight lines make 180°.

Let’s focus on the point where the 140°, 40°, and 120° meet.

Actually, notice that 140° and 40° are adjacent on a straight line? Let’s see:

If you look at the left side, there’s a straight line going from bottom left to top right. On that line, we have an angle of 140° and next to it 40° — do they form a straight line?

140° + 40° = 180° → Yes! So those two angles are on a straight line. That means the line segment between them is straight.

Now, at the intersection point inside the figure, we have angles: 40°, 120°, and others.

Actually, let’s consider the triangle that includes the 120° angle and the 15° angle.

Wait — here’s a better way.

Let’s label some points mentally.

Imagine the big triangle formed by the bottom horizontal line, the left slanted line, and the right slanted line.

But actually, the figure looks like two triangles sharing a common vertex or overlapping.

Alternatively, let’s use the concept of exterior angles or triangle angle sums carefully.

Let me try this:

Consider the small triangle at the top: angles 20° and 40°, so third angle is 120° — which matches the labeled 120° angle. So that 120° is the angle at the intersection point, inside that small triangle.

Now, that 120° angle and the angle opposite to it (vertically opposite) should be equal — but in the diagram, the 120° is already shown, and below it, there’s another angle.

Actually, at the intersection point, four angles are formed. We know one is 120°, and since vertically opposite angles are equal, the angle opposite to it is also 120°.

Then the other two angles must add up to 360° - 120° - 120° = 120°, and since they are vertically opposite, each is 60°.

Wait — is that correct?

No: when two lines intersect, they form two pairs of vertically opposite angles. If one angle is 120°, then its vertical opposite is also 120°, and the other two angles (adjacent) must each be 60°, because 120° + 60° = 180° on a straight line.

Yes! So at the intersection point, the angles are: 120°, 60°, 120°, 60°.

Now, look at the triangle on the bottom right: it has angles 15°, and one of these 60° angles (the one adjacent to the 120°), and the third angle is part of x°.

Actually, the angle inside that small triangle at the intersection point is 60° (since it's adjacent to 120° on a straight line).

So in the bottom-right triangle: angles are 15°, 60°, and the third angle (let’s call it B).

Sum: 15° + 60° + B = 180° → B = 105°

But B is not x — x is the angle at the bottom, on the straight line.

Wait — x is the angle at the bottom vertex, between the bottom horizontal line and the slanted line going up.

Actually, looking back at the diagram description: x° is at the bottom, on the left side of the bottom horizontal line? No — typically in such diagrams, x is the angle at the bottom vertex of the large triangle.

Perhaps I need to think differently.

Let me consider the entire large triangle.

The large triangle has vertices: bottom left, bottom right, and top.

At the bottom right, angle is 15°.

At the top, what is the angle? From the top triangle, we had 20° and 40°, so the top angle of the large triangle might be 20° + something.

Actually, the 20° is at the very top, and the 40° is below it, so the total top angle of the large triangle is 20° + 40° = 60°? But that doesn't seem right because the 40° is not at the vertex.

I think I'm overcomplicating.

Let me try using the fact that the sum of angles in any triangle is 180°, and look for a triangle that includes x.

Notice that x is at the bottom, and it's part of a triangle that also includes the 140° angle and the 120° angle? Not directly.

Another idea: the angle labeled 140° is on the left side. Since it's on a straight line with the 40° angle (as 140+40=180), that means the line is straight.

Now, at the point where the 140° angle is, it's also part of a triangle.

Let’s consider the triangle that has angles: 140°, and then the angle at the bottom left, and the angle at the intersection.

This is messy.

Let me search for a standard method for such problems.

I recall that in such figures, sometimes we can use the "angle sum around a point" or "exterior angle theorem".

Let’s try the exterior angle theorem.

For example, in the top triangle with angles 20° and 40°, the exterior angle at the bottom would be 20° + 40° = 60°.

And that exterior angle might be equal to the angle inside the other triangle.

At the intersection point, if the exterior angle from the top triangle is 60°, then the vertically opposite angle is also 60°, and the adjacent angle is 120°, which matches what we have.

Now, in the bottom-right triangle, we have angles: 15°, and the 60° angle (from the intersection), so the third angle is 180° - 15° - 60° = 105°.

This 105° angle is at the bottom right vertex of that small triangle, but it's not x.

x is at the bottom left of the large figure.

Let’s consider the large triangle that has vertices at: bottom left, bottom right, and top.

At bottom right, the angle is 15°.

At the top, what is the angle? The top angle is composed of the 20° and the angle from the 40° triangle, but actually, the 20° is the top angle of the small triangle, and the 40° is at the side.

Perhaps the large triangle's top angle is 20° + the angle between the 40° and the top, but it's complicated.

Another approach: let's calculate the angle at the bottom left of the large triangle.

From the left side, we have the 140° angle. This 140° is the exterior angle for the triangle at the bottom left.

In the triangle that has the x° angle at the bottom, and the 140° angle is adjacent to it on the straight line.

Since 140° and the angle inside the triangle at that vertex are on a straight line, they add to 180°.

So, the interior angle at the bottom left vertex is 180° - 140° = 40°.

Is that right? Let's see: the 140° is marked outside, so yes, the interior angle of the large triangle at the bottom left is 180° - 140° = 40°.

Similarly, at the bottom right, we have 15°, which is already the interior angle.

Now, what about the top angle of the large triangle?

The top angle is made up of the 20° and the angle from the other part.

From the top small triangle, we have angles 20° and 40°, so the third angle is 120°, as before.

This 120° angle is at the intersection point, and it is also the angle between the two lines.

For the large triangle, the top angle is the angle at the very top, which is 20°.

Is that correct? In the diagram, the 20° is at the apex, so yes, the top angle of the large triangle is 20°.

But then, if the large triangle has angles: bottom left 40°, bottom right 15°, top 20°, sum is 40+15+20=75°, which is not 180° — impossible.

So my assumption is wrong.

The top angle of the large triangle is not just 20°; it must include more.

Let's think: the large triangle's top vertex is the same as the small triangle's top vertex, so the top angle is 20°.

But then the sum is too small, so perhaps the large triangle is not the one I'm thinking.

Perhaps the large triangle is formed by the bottom line, the left line, and the right line, but the right line is not straight; it's bent.

I think I need to accept that and use a different strategy.

Let me consider the triangle that includes x, the 140° angle, and the 120° angle.

At the point where the 140° and 120° meet, they are not in the same triangle.

Let's list all the angles we know.

From the top: in the small triangle, angles are 20°, 40°, and 120° (at the intersection).

At the intersection point, the vertically opposite angle to 120° is 120°, and the adjacent angles are 60° each.

Now, consider the triangle on the bottom left: it has the x° angle at the bottom, the 140° angle at the left, and the angle at the intersection.

The angle at the intersection for this triangle is the one adjacent to the 120°, which is 60° (since 120° + 60° = 180° on the straight line).

So in this bottom-left triangle, angles are: x°, 140°, and 60°? But 140° is not an interior angle; it's the exterior.

Let's clarify.

The 140° angle is marked on the outside, so for the triangle, the interior angle at that vertex is 180° - 140° = 40°.

Similarly, at the intersection point, for the bottom-left triangle, the interior angle is the 60° angle (because the 120° is on the other side).

So in the bottom-left triangle, angles are:
- At bottom left: x° (this is the angle we want, but is it the interior angle? In the diagram, x° is likely the interior angle at the bottom vertex.
- At the left vertex: interior angle is 180° - 140° = 40°
- At the intersection point: 60° (as established)

Sum of angles in this triangle: x + 40° + 60° = 180°
So x + 100° = 180°
Therefore, x = 80°

Is that it? Let me verify.

If x = 80°, then in the bottom-left triangle: 80° + 40° + 60° = 180° — good.

Now, let's check the other parts to ensure consistency.

In the bottom-right triangle: angles at intersection is 60° (the other 60°), at bottom right is 15°, so third angle is 180° - 60° - 15° = 105°.

This 105° angle is at the bottom right vertex of that small triangle, but in the large figure, at the bottom right, the angle is 15°, which is already given, so it should be fine.

Also, at the top, the small triangle has 20° + 40° + 120° = 180° — good.

And the angles at the intersection: 120°, 60°, 120°, 60° — sum to 360° — good.

So everything checks out.

Therefore, x = 80°.

Final Answer: 80