SWBAT apply & identify the equilateral and isosceles triangle ... - Free Printable

Let’s solve each problem one by one. We’ll use triangle properties:
- In any triangle, the sum of interior angles is 180°.
- If two sides are equal (marked with same number of ticks), then the angles opposite them are equal → Isosceles Triangle.
- If all three sides are equal → Equilateral Triangle (all angles = 60°).
- Right triangles have one 90° angle.

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Problem 13:
Triangle with sides: 6x+3, 8x–21, 7x–10
Two base angles marked equal → so it’s isosceles → two sides equal.
Which two? The ones adjacent to the equal angles → left and right sides:
So set 6x + 3 = 8x – 21
→ 3 + 21 = 8x – 6x
→ 24 = 2x → x = 12

Check third side: 7x – 10 = 7(12) – 10 = 84 – 10 = 74
Left side: 6(12)+3=75, Right side: 8(12)-21=96-21=75 → yes, equal.
But wait — if two sides are equal, angles opposite are equal — that matches the diagram.
So x = 12.

Wait — but let’s check if maybe the base is the unequal side? Actually, in the diagram, the two bottom angles are marked equal → so the two legs (left and right) are equal → correct.

✔ Final Answer for 13: x = 12

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Problem 14:
Triangle with angles: top angle not labeled, left angle = 3x–15, right angle = x–7, and side labels: left side = 2x+10, right side = ? Wait — actually, looking again:

Actually, the markings: two angles at the base are marked with arcs → meaning they are equal? No — wait, in problem 14, the two *sides* are marked with single tick marks → so those two sides are equal → so the angles opposite them are equal.

The two sides with ticks are the left and right sides → so angles opposite them are the bottom-left and bottom-right angles? Wait no — in a triangle, side opposite an angle.

Actually, let's label:

Top vertex: angle unknown
Bottom-left vertex: angle = 3x – 15
Bottom-right vertex: angle = x – 7
Side opposite bottom-left angle is the right side → labeled “x – 7”? That doesn’t make sense.

Wait — I think I misread. Let me re-express:

In problem 14:
It’s a triangle pointing down.
Top side: labeled “2x + 10”
Left side: labeled “3x – 15”
Right side: labeled “x – 7”
And there are tick marks on the left and right sides → meaning left side = right side.

So: 3x – 15 = x – 7
→ 3x – x = –7 + 15
→ 2x = 8 → x = 4

Check: left side = 3(4)–15 = 12–15 = –3 → negative length? Impossible!

That can’t be. So maybe the tick marks are on the angles? Looking back at original image description — user said “two angles marked with arcs” — but in text we don’t see arcs.

Wait — perhaps in problem 14, the two *angles* at the base are marked as equal? Because in many such problems, when two angles are marked with same arc, they’re equal.

Assume that: the two bottom angles are equal → so 3x – 15 = x – 7 → same equation → x=4 → gives negative side? Not possible.

Alternative: maybe the two sides with ticks are the top and one other? But top is labeled 2x+10, and no tick mentioned on it.

Perhaps the tick marks are on the two equal angles — which would mean the sides opposite are equal.

If the two base angles are equal, then the two legs (left and right sides) are equal → again 3x–15 = x–7 → x=4 → invalid.

Unless... maybe the expression is for the angles, not sides? Let me reread the problem statement from user input:

“14. [triangle] with labels: top side 2x+10, left side 3x–15, right side x–7” — and probably the two base angles are marked equal? Or the two sides?

Given that x=4 gives negative length, likely the expressions are for ANGLES, not sides.

Ah! That makes more sense. In many worksheets, when you see expressions like "3x–15" inside the angle, it’s the measure of the angle.

Looking back at problem 13: “6x+3”, “8x–21”, “7x–10” — these were likely side lengths because we set two equal based on angle marks.

But in problem 14, if the numbers are inside the angles, then:

Assume:
Angle at bottom-left = 3x – 15
Angle at bottom-right = x – 7
And since the two sides are marked equal (ticks on left and right sides), then the angles opposite them are equal.

Side opposite bottom-left angle is the right side → but we don't have its angle label.

This is confusing without seeing the image clearly.

Alternative approach: in problem 14, if the two base angles are marked with arcs (equal), then:

Set 3x – 15 = x – 7 → x=4 → then angles are 3(4)-15= -3 → impossible.

So perhaps the expressions are for the sides, and the tick marks indicate which sides are equal.

Maybe the top side and one leg are equal? But only left and right have ticks.

Another idea: perhaps "2x+10" is the top angle? No, usually sides are labeled on sides.

Let me try assuming that the two angles at the base are equal, and the expressions given are for the sides — but that doesn't help directly.

Perhaps for problem 14, the triangle has two sides equal (left and right), so their opposite angles are equal.

The angle opposite the left side is the bottom-right angle, and vice versa.

So if left side = right side, then angle opposite left side = angle opposite right side.

Angle opposite left side is the bottom-right angle, which is labeled "x – 7"? And angle opposite right side is bottom-left angle, labeled "3x – 15"?

So set 3x – 15 = x – 7 → same as before → x=4 → still negative.

This suggests that either the expressions are for angles, or there's a mistake.

Let's calculate what x should be for positive lengths.

Suppose left side = 3x–15 >0 → x>5
Right side = x–7 >0 → x>7
Top side = 2x+10 >0 always for x>0.

If left = right, 3x–15 = x–7 → x=4 <7 → invalid.

If top = left: 2x+10 = 3x–15 → 10+15 = 3x–2x → x=25
Then left side = 3(25)–15=75–15=60
Right side = 25–7=18
Top = 2(25)+10=60 → so top = left =60, right=18 → isosceles with top and left equal.

But are there tick marks on top and left? The user didn't specify, but in the initial description, for problem 14, it says "two sides marked with ticks" — probably left and right, but that leads to contradiction.

Perhaps in the image, the tick marks are on the two equal angles, not sides.

Let me assume that the two base angles are equal, and the expressions "3x–15" and "x–7" are the measures of those angles.

Then 3x – 15 = x – 7 → x=4 → angles = 3(4)-15= -3 — impossible.

Unless it's 3x+15 or something, but it's written as 3x–15.

Another possibility: the expression "2x+10" is the vertex angle, and the two base angles are equal, so each is (180 - (2x+10))/2, but we have expressions for them.

I think there might be a misinterpretation.

Let's look at problem 15 for context.

Problem 15:
Right triangle, one angle 50°, another angle y, and the third angle is x, with a square at the corner indicating 90°.

So angles: 90°, 50°, and y? But it says "x" and "y" — probably x is at the top, y at the bottom-left.

Sum: x + y + 50 = 180, and since it's right-angled, one angle is 90°.

The right angle is at the bottom-left, marked with square, so that angle is 90° — but it's labeled "y"? That can't be.

In the description: "15. [right triangle] with angles: top=x, bottom-left=y, bottom-right=50°, and right angle mark at bottom-left."

If there's a right angle mark at bottom-left, then y = 90°.

Then x + 90 + 50 = 180 → x = 40°.

But why label it y if it's 90°? Perhaps y is not the right angle.

Re-read: "15. [triangle] with x at top, y at bottom-left, 50° at bottom-right, and a right angle symbol at the bottom-left corner."

So the right angle is at bottom-left, so y = 90°.

Then x + 90 + 50 = 180 → x = 40°.

So for 15: x=40, y=90.

But typically, if it's marked as right angle, we know it's 90, so y=90, x=40.

Okay, that works.

Back to 14.

Perhaps in 14, the expressions are for the angles, and the tick marks indicate which angles are equal.

Suppose the two base angles are marked with arcs, so they are equal, and their measures are 3x–15 and x–7.

Then 3x–15 = x–7 → x=4, but then angle = 3(4)-15= -3 — impossible.

Unless it's 3x+15, but it's written as 3x–15.

Perhaps "3x–15" is the measure of the top angle, and the two base angles are equal.

Let me try that.

Assume top angle = 2x+10 (since it's on the top side, but usually sides are labeled on sides, angles inside).

I think I need to assume that for problem 14, the two sides with tick marks are equal, and the expressions are for the sides, and despite x=4 giving negative, perhaps it's a typo, or perhaps I have the wrong pair.

Another idea: perhaps the tick marks are on the two equal angles, and the expressions are for the sides, but that doesn't help.

Let's calculate the sum of angles if we assume the expressions are for angles.

Suppose the three angles are: A = 2x+10, B = 3x–15, C = x–7.

Sum = 2x+10 + 3x–15 + x–7 = 6x –12 = 180 → 6x = 192 → x=32.

Then angles: 2(32)+10=74, 3(32)-15=96-15=81, 32-7=25. Sum 74+81+25=180 — good.

Now, are any two angles equal? 74,81,25 — no, but in the diagram, there are tick marks on two sides, which would imply two angles equal, but here no two angles are equal.

Unless the tick marks are on the sides to indicate which are equal, but with x=32, sides would be proportional to angles, but we don't have side lengths.

Perhaps for problem 14, the tick marks are on the two sides that are equal, and we need to set those expressions equal.

But earlier, setting left=right gave x=4, invalid.

Setting top=left: 2x+10 = 3x–15 → x=25, as before.

Then sides: top=2(25)+10=60, left=3(25)-15=60, right=25-7=18.

So two sides equal: top and left.

If in the diagram, the top and left sides have tick marks, then it's correct.

Perhaps the user meant that the two sides with ticks are the ones that are equal, and in this case, it's top and left.

Or perhaps in the image, the ticks are on the top and left sides.

Since x=25 gives positive lengths, and sum of angles will work, let's go with that.

For now, I'll assume that for problem 14, the two equal sides are the top and left, so 2x+10 = 3x–15 → x=25.

Then the missing value is x=25.

But the problem asks for "missing values", and in this case, only x is to be found.

So for 14: x=25.

Let's move on and come back if needed.

Problem 15: As above, right angle at bottom-left, so y=90°, then x + 90 + 50 = 180 → x=40°.

So x=40, y=90.

Problem 16:
Triangle with two sides marked with ticks → isosceles.
Angles: bottom-left = 2x, bottom-right = 5x–30, and since two sides equal, the base angles are equal.

Which sides are equal? The two legs, so the base angles are the bottom ones.

So set 2x = 5x – 30
→ 30 = 5x – 2x = 3x → x=10

Check: 2x=20, 5x-30=50-30=20 — equal, good.

Vertex angle = 180 - 20 - 20 = 140°, not asked.

So x=10.

Problem 17:
Right triangle, with right angle at bottom-right (marked with square), and one acute angle 45°, so it's isosceles right triangle.

Sides: hypotenuse = x, one leg = y, other leg = ? Not labeled, but since 45-45-90, legs are equal.

The angle at bottom-left is 45°, so the two legs are equal.

The leg adjacent to 45° is y, and the other leg is not labeled, but since it's isosceles, both legs equal.

Hypotenuse is x.

In 45-45-90 triangle, legs are equal, hypotenuse = leg * √2.

But here, we have expressions: one leg is y, hypotenuse is x, and the other leg is not given, but since it's isosceles, the two legs are equal, so the unlabeled leg is also y.

But we have only x and y, and no other information.

The angle is 45°, so tan(45°) = opposite/adjacent = y / y = 1, which is true, but doesn't give relation between x and y.

We need another equation.

Perhaps the side labeled "x" is the hypotenuse, and "y" is a leg, and since it's 45-45-90, x = y√2.

But we have only one equation, two variables.

Unless there's more.

In the diagram, there are tick marks on the two legs? The user said "two sides marked with double ticks" — in problem 17: "x" on hypotenuse, "y" on one leg, and the other leg has double ticks, and the first leg has single tick? Let's see.

From user input: "17. [triangle] with x on hypotenuse, y on one leg, and the other leg has double ticks, and the angle at bottom-left is 45°."

Also, there is a right angle at bottom-right.

So, the two legs: one is labeled y, the other has double ticks — but no label.

Probably, the double ticks indicate that the two legs are equal, so the unlabeled leg is equal to y.

So both legs are y, hypotenuse is x.

Then by Pythagoras: y^2 + y^2 = x^2 → 2y^2 = x^2 → x = y√2.

But we have two variables, no numerical value.

Perhaps we need to find x and y, but there's no additional info.

Unless the "45°" is used, but in isosceles right triangle, it's automatic.

Perhaps the expression is for the sides, and we need to use the angle.

Another thought: perhaps "x" and "y" are to be found, but we need another condition.

Maybe the double ticks mean that the two legs are equal, so we can set up, but still.

Perhaps in the diagram, the side with double ticks is labeled, but in text, it's not.

Let's assume that the two legs are equal, so if one leg is y, the other is also y, hypotenuse x, so x = y√2.

But that's not a numerical answer.

Perhaps the 45° is redundant, and we need to express, but the problem asks for "missing values", implying numerical.

Perhaps I missed something.

Another idea: perhaps "x" is not the hypotenuse, but let's see the description: "x" on the hypotenuse, "y" on one leg.

Perhaps the angle 45° is at the top, but user said "at bottom-left".

Let's calculate the angles.

Right angle at bottom-right, 45° at bottom-left, so top angle is 45°, so yes, isosceles, legs equal.

So if leg is y, hypotenuse x, then x = y * √2.

But we can't find numerical values without more info.

Unless the "double ticks" indicate that the two legs are equal, and perhaps there's a length given, but no.

Perhaps in the diagram, the side with double ticks is labeled with a number, but in text, it's not specified.

This is problematic.

Perhaps for problem 17, the "x" and "y" are to be related, but the problem likely expects numerical values, so perhaps I have a mistake.

Let's look at problem 18.

Problem 18:
Right triangle, right angle at bottom-left.
Sides: left leg = 3x+1, bottom leg = 4x–20, and there are tick marks on the two legs? User said "tick marks on the two legs" — so left leg = bottom leg.

So 3x+1 = 4x–20
→ 1 + 20 = 4x – 3x → x=21

Then left leg = 3(21)+1=63+1=64, bottom leg = 4(21)-20=84-20=64, good.

Hypotenuse not asked.

So x=21.

Problem 19:
Triangle with two sides marked with ticks → isosceles.
Angles: bottom-left = 28°, top = x, bottom-right = y.

Since two sides equal, the base angles are equal.

Which sides are equal? The two legs, so the base angles are the bottom ones.

So bottom-left = bottom-right = 28°? But then y=28°, and x = 180 - 28 - 28 = 124°.

But is that correct? The two sides with ticks are the left and right sides, so yes, the angles opposite them are the bottom-right and bottom-left, so if left side = right side, then angle opposite left side = angle opposite right side.

Angle opposite left side is bottom-right angle, angle opposite right side is bottom-left angle.

So if left side = right side, then bottom-right angle = bottom-left angle.

Given bottom-left = 28°, so bottom-right = y = 28°.

Then x = 180 - 28 - 28 = 124°.

So x=124, y=28.

Problem 20:
Triangle with two angles marked with arcs → so those two angles are equal.
Expressions: left angle = 3x–6, bottom angle = x+10, and the third angle is not labeled.

Since two angles are equal, and likely the two marked ones are equal.

So set 3x–6 = x+10
→ 3x – x = 10 + 6 → 2x = 16 → x=8

Then angles: 3(8)-6=24-6=18, x+10=8+10=18, so both 18°, then third angle = 180-18-18=144°.

Good.

So x=8.

Problem 21:
Triangle with a line from top to base, creating two smaller triangles.
Left small triangle has angles: at top 54°, at bottom-left not labeled, at the new point not labeled.
There are tick marks: on the left side of big triangle, and on the segment from top to base, and on the bottom-right part.

Specifically: the left side of big triangle has a tick, the segment from top to base has a tick, and the bottom-right side has a tick.

Also, the angle at top is 54°, and at bottom-right is x.

Since the left side and the segment from top to base have ticks, they are equal, so the triangle formed by top, bottom-left, and the foot is isosceles with those two sides equal.

So in the left small triangle, sides: left side = segment, so angles opposite are equal.

Angle opposite left side is the angle at the foot (on the base), angle opposite the segment is the angle at bottom-left.

So let's call the foot D on base BC, with B bottom-left, C bottom-right, A top.

So AD is the segment.

Given AB = AD (ticks), so in triangle ABD, AB = AD, so angles opposite equal: angle at D = angle at B.

Angle at A in triangle ABD is part of the 54°.

The 54° is the angle at A for the big triangle, but in triangle ABD, the angle at A is the same 54°? No, because AD is inside, so the 54° is split.

The user said "54°" at the top, and it's the angle of the big triangle, so angle BAC = 54°.

But with AD drawn, angle BAD and angle CAD are parts of it.

In the diagram, the 54° is likely the angle at A for the left small triangle, or for the big one.

From description: "54°" at the top, and it's probably the angle between AB and AC, but with AD, it's divided.

Perhaps the 54° is the angle at A in triangle ABD.

Assume that in triangle ABD, angle at A is 54°, and AB = AD, so it's isosceles with AB=AD, so base angles equal: angle at B = angle at D.

Sum of angles in triangle ABD: 54 + angle B + angle D = 180, and angle B = angle D, so 2*angle B = 126, angle B = 63°.

So angle at B is 63°.

Now, this is the bottom-left angle of the big triangle.

Now, the big triangle has angles: at B 63°, at A 54°, so at C = 180 - 63 - 54 = 63°.

Oh! So angle at C is also 63°, so big triangle is isosceles with AB = AC? But we have ticks on AB, AD, and DC.

Angle at B = 63°, angle at C = 63°, so yes, AB = AC.

But in the diagram, there is a tick on DC, and on AB and AD.

Angle at C is x, and we have 63°, so x=63°.

But let's confirm.

From above, in triangle ABD, AB=AD, angle at A=54°, so angles at B and D are (180-54)/2 = 63° each.

So angle ABC = 63°.

Now, angle BAC = 54°, so in big triangle ABC, angle ACB = 180 - 54 - 63 = 63°.

So x = angle at C = 63°.

Additionally, there is a tick on DC, and since angle at C is 63°, and if DC is equal to something, but not necessary for x.

So x=63.

Problem 22:
Two triangles sharing a side, with various tick marks.

Left triangle: sides with ticks: one side has single tick, another has double ticks.

Right triangle: sides with ticks: one has single, one has double, and an angle 2x°.

Also, left triangle has angle 5x°.

Probably, the tick marks indicate equal sides.

Assume that the shared side is common.

In left triangle, suppose sides: let's say side AB has single tick, side BC has double ticks, etc.

Perhaps the single tick means those sides are equal across or within.

Typically, same number of ticks mean equal length.

So in left triangle, if two sides have the same number of ticks, they are equal.

Here, left triangle has one side with single tick, one with double ticks — so probably not equal within.

Perhaps the single tick on left triangle corresponds to single tick on right triangle, meaning those sides are equal.

Similarly for double ticks.

Also, there is an angle 5x° in left triangle, 2x° in right triangle.

And they share a side.

Perhaps the shared side has no tick, or has a tick.

To simplify, assume that the side with single tick in left triangle is equal to the side with single tick in right triangle, and similarly for double ticks.

But we need to see which sides.

Perhaps the two triangles are congruent or something.

Another idea: perhaps the angle 5x° and 2x° are related.

Let's denote the shared side as S.

In left triangle, suppose the side opposite to 5x° is something.

Perhaps use the fact that the sum of angles around a point or something.

Notice that the two triangles together form a quadrilateral or something, but not specified.

Perhaps the angles at the shared vertex are supplementary or something.

Assume that the two triangles are on a straight line or something.

Perhaps the angle between them is 180°.

Let's look for standard configuration.

Perhaps the 5x° and 2x° are vertical angles or adjacent.

Another thought: in the diagram, the two triangles share a vertex, and the angles 5x° and 2x° are at that vertex, and they are adjacent, forming a straight line or something.

Suppose that at the shared vertex, the two angles 5x° and 2x° are adjacent and form a straight line, so 5x + 2x = 180 → 7x=180 → x=180/7≈25.71, not nice.

Perhaps they are vertical angles, but then equal, so 5x=2x → x=0, impossible.

Perhaps the angles are in different places.

Let's assume that the side with single tick in left triangle is equal to the side with single tick in right triangle, and similarly for double ticks, and the shared side is common, so the two triangles have two sides equal and included angle or something.

But we don't know the included angle.

Perhaps the angle 5x° is in left triangle, 2x° in right, and they are corresponding.

Another idea: perhaps the two triangles are isosceles themselves.

For example, in left triangle, if two sides have the same tick, but here different ticks.

Unless the single tick and double tick are for different pairs.

Perhaps for the left triangle, the two sides with ticks are not equal, but the tick indicates comparison with other triangle.

I recall that in some problems, when two triangles have sides with matching ticks, those sides are equal.

So suppose that the side with single tick in left triangle is equal to the side with single tick in right triangle.

Similarly, the side with double ticks in left is equal to the side with double ticks in right.

And the shared side is common to both.

So the two triangles have three sides equal: shared side, single-tick side, double-tick side, so they are congruent by SSS.

Then corresponding angles are equal.

So the angle 5x° in left triangle corresponds to some angle in right triangle.

Depending on which angle.

If the 5x° is opposite the single-tick side in left, and in right, the angle opposite the single-tick side is 2x°, then 5x = 2x, impossible.

Perhaps 5x° is at the vertex where the single-tick and shared side meet, etc.

Perhaps the angle 5x° and 2x° are the angles at the shared vertex.

Suppose that at the shared vertex, the angle in left triangle is 5x°, in right triangle is 2x°, and they are adjacent, so if they form a straight line, 5x + 2x = 180, x=180/7, not integer.

If they are on the same side, but usually not.

Another possibility: the two angles are vertical angles, but then equal, 5x=2x, x=0.

Perhaps the 5x° and 2x° are not at the shared vertex.

Let's assume that the two triangles are congruent, so all corresponding angles equal.

Then the angle 5x° in left must equal some angle in right, say the 2x° or another.

But there is only one angle labeled in each.

Perhaps the 2x° is the corresponding angle to 5x°, so 5x = 2x, impossible.

Unless it's not corresponding.

Perhaps the sum of angles.

Let's calculate the third angle.

In left triangle, angles sum to 180, but we have only one angle 5x°, so not enough.

Perhaps the tick marks indicate that the triangles are isosceles.

For example, in left triangle, if two sides have the same number of ticks, but here one has single, one has double, so probably not.

Unless the shared side has a tick, but not specified.

Perhaps the side with single tick in left is equal to the shared side, but no tick on shared side.

I think I need to guess that the two angles are related by the geometry.

Another idea: perhaps the two triangles share the side, and the angles 5x° and 2x° are at the ends, and the figure is symmetric or something.

Perhaps the line is straight, so the sum of angles on one side is 180°.

Suppose that at the shared vertex, the angle from left triangle is 5x°, from right triangle is 2x°, and they are on a straight line, so 5x + 2x = 180, x=180/7, but let's see if it makes sense.

Then in left triangle, angles: 5x = 900/7 ≈128.57°, and other angles.

But we have tick marks, so perhaps isosceles.

Suppose in left triangle, the two sides with ticks are equal, but they have different ticks, so unlikely.

Perhaps the single tick means that side is equal to the single-tick side in the other triangle, etc.

Let's assume that the side with single tick in left is equal to the side with single tick in right, and the side with double ticks in left is equal to the side with double ticks in right, and the shared side is common, so SSS congruence.

Then the angles are equal.

So the angle 5x° in left triangle corresponds to the angle in right triangle that is in the same position.

If the 2x° is the corresponding angle, then 5x = 2x, impossible.

If the 2x° is not the corresponding angle, but another angle, then we don't know.

Perhaps the 5x° and 2x° are the apex angles or something.

Another thought: perhaps the two triangles are on a base, and the angles at the base are given.

Let's look for the answer.

Perhaps 5x and 2x are supplementary because they are adjacent on a straight line.

So 5x + 2x = 180, x=180/7, but let's calculate numerical.

180/7 ≈25.714, then 5x≈128.57, 2x≈51.43, sum 180, good for straight line.

Then in each triangle, we can find other angles, but the problem may just want x, and it's valid.

But usually answers are integer, so perhaps not.

Perhaps the angles are vertical, but then equal.

Another idea: perhaps the 5x° and 2x° are the same angle, but that doesn't make sense.

Or perhaps it's a typo, and it's 5x and 5x or something.

Let's assume that the two triangles are identical, so 5x = 2x, impossible.

Perhaps the angle in the right triangle is 2x, but it's the base angle, while 5x is apex.

I recall that in some problems, the angle between the two triangles is 180 degrees.

Perhaps the sum of the two angles is 180 if they are adjacent on a straight line.

And in many textbooks, they use that.

So let's go with 5x + 2x = 180, so 7x=180, x=180/7.

But let's see if it fits with tick marks.

Perhaps the tick marks indicate that the sides are equal, and with x=180/7, it works, but not nice.

Another possibility: perhaps the 5x° and 2x° are not the angles at the shared vertex, but at other vertices, and the shared side is between them.

Perhaps use the law of sines, but too advanced.

Let's try to assume that the two triangles have the shared side, and the sides with single ticks are equal, say length a, sides with double ticks are equal, say length b, shared side c.

Then in left triangle, sides a,b,c, with angle 5x° opposite to say side a.

In right triangle, sides a,b,c, with angle 2x° opposite to side a or b.

If in left, angle 5x° is opposite side a, in right, angle 2x° is opposite side a, then by law of sines, in left: a / sin(5x) = 2R, in right: a / sin(2x) = 2R, so sin(5x) = sin(2x), so 5x = 2x or 5x = 180-2x.

5x=2x => x=0, invalid.

5x = 180 - 2x => 7x=180, x=180/7 again.

Same as before.

If in right, angle 2x° is opposite side b, then a / sin(5x) = b / sin(angle opposite b in left), etc, complicated.

So probably x=180/7.

But let's keep it as fraction.

Perhaps the angle 2x° is the whole angle at that vertex, but in the triangle, it's part.

I think for now, I'll take x=180/7 for problem 22.

But let's move to others and come back.

Problem 23:
Triangle with two sides marked with ticks → isosceles.
Angles: left = 3x+10, right = 5x–10, and since two sides equal, the base angles are equal.

Which sides are equal? The two legs, so the base angles are the bottom ones, but here the angles are at the base? The expressions are for the angles at the base.

So if the two base angles are equal, then 3x+10 = 5x–10
→ 10+10 = 5x–3x → 20 = 2x → x=10

Then angles: 3(10)+10=40, 5(10)-10=40, so both 40°, vertex angle = 180-40-40=100°.

Good.

So x=10.

Problem 24:
Right triangle, right angle at bottom-left.
Sides: left leg = x, bottom leg = ? not labeled, but there are tick marks: on the left leg and on the bottom leg? User said "tick marks on the two legs" — so left leg = bottom leg.

So it's isosceles right triangle.

Then the two acute angles are 45° each.

The angle at top is x? In the diagram, "x" is at the top angle.

So x = 45°.

Because in isosceles right triangle, acute angles are 45°.

So x=45.

Problem 25:
Quadrilateral or two triangles? From description: "4x+2" on left side, "6x–30" on right side, angle 60° at top, and right angle at bottom-left.

Also, there are tick marks: on the left side and on the diagonal or something.

Specifically: "4x+2" on left side, "6x–30" on right side, 60° at top, right angle at bottom-left, and tick marks on the left side and on the segment from top to bottom-right or something.

Probably, it's a kite or something.

Assume that the two triangles share the diagonal.

Left triangle: sides 4x+2, and the diagonal, and bottom.

Right triangle: sides 6x–30, diagonal, and bottom.

Tick marks: on the left side (4x+2) and on the right side (6x–30)? But they are different.

User said: "tick marks on the left side and on the diagonal" or something.

From input: "25. [figure] with 4x+2 on left, 6x-30 on right, 60° at top, right angle at bottom-left, and tick marks on the left side and on the segment from top to bottom-right."

Also, there is a right angle at bottom-left, so in the left triangle, angle at bottom-left is 90°.

And 60° at top, so in the left triangle, angles: at top 60°, at bottom-left 90°, so at bottom-right of left triangle is 30°.

Then the side opposite 30° is half the hypotenuse, but let's see.

In left triangle, angles 90°, 60°, 30°, so sides in ratio 1 : √3 : 2.

Specifically, side opposite 30° is shortest.

Here, the side adjacent to 90° and 60° is the bottom side, etc.

Label: let A be top, B bottom-left, C bottom-right.

So triangle ABC, with right angle at B, angle at A is 60°, so angle at C is 30°.

Side AB = 4x+2 (left side), side BC = ? , side AC = diagonal.

In triangle ABC, angle at A 60°, at B 90°, at C 30°.

Side opposite to A is BC, opposite to B is AC, opposite to C is AB.

Standard: in 30-60-90, side opposite 30° is half hypotenuse.

Here, angle at C is 30°, so side opposite to it is AB.

So AB = (1/2) * hypotenuse AC.

Hypotenuse is AC, since opposite right angle at B.

So AB = (1/2) * AC.

But AB = 4x+2, so AC = 2*(4x+2) = 8x+4.

Now, there is another triangle: probably triangle ADC or something, but user said "6x-30" on the right side, which is likely AC or BC.

The right side is probably AC or the other side.

User said "6x-30" on the right side, and there is a tick mark on the left side and on the diagonal.

Also, there is a right angle at bottom-left, which is B, and probably the figure is quadrilateral ABCD or something, but likely it's triangle ABC with D on BC or something.

Perhaps it's a single triangle, but with a point.

Another interpretation: perhaps the "right side" is the side from A to C, and "left side" is from A to B, and there is a point D on BC, but complicated.

Perhaps the figure is a triangle with a median or something.

Let's read: "25. [figure] with 4x+2 on left, 6x-30 on right, 60° at top, right angle at bottom-left, and tick marks on the left side and on the segment from top to bottom-right."

Also, "6x-30" is on the right side, which might be the side from top to bottom-right, i.e., AC.

And "4x+2" on left side, AB.

And tick marks on AB and on AC? But then AB = AC, but in our calculation, AB = 4x+2, AC = 8x+4, so 4x+2 = 8x+4 → 2-4 = 8x-4x → -2 = 4x → x= -0.5, invalid.

Perhaps the tick marks are on AB and on BC or something.

Another idea: perhaps the "segment from top to bottom-right" is AC, and it has a tick, and AB has a tick, so AB = AC.

But as above, in the left triangle, if AB = AC, but angle at B is 90°, angle at A is 60°, then if AB = AC, but AC is hypotenuse, AB is leg, so AB < AC, cannot be equal unless degenerate.

So not.

Perhaps the right angle is not in the left triangle alone.

Perhaps the figure is a quadrilateral with diagonal.

Assume that there is a diagonal from A to C, and B and D on the base.

But user mentioned "bottom-left" and "bottom-right", so probably points B and C on base, A top.

With right angle at B, so AB perpendicular to BC.

Angle at A is 60°, so in triangle ABC, angle at A is 60°, at B 90°, so at C 30°.

Then side AB = 4x+2, side BC = ? , side AC = ?.

As above, AB opposite 30°? No.

In triangle ABC:
- angle at B: 90°
- angle at A: 60°
- angle at C: 30°

Side opposite angle A (60°) is BC.
Side opposite angle B (90°) is AC (hypotenuse).
Side opposite angle C (30°) is AB.

So AB = opposite 30° = (1/2) * hypotenuse AC.

So AB = (1/2) * AC.

So 4x+2 = (1/