Solving Problems Using Angle Relationships #1 | Interactive ... - Free Printable

Let's solve each problem step by step using angle relationships. We'll use properties of angles such as:

- Complementary angles: Add up to 90°
- Supplementary angles: Add up to 180°
- Vertical angles: Opposite angles formed by intersecting lines are equal
- Angles around a point: Sum to 360°
- Linear pairs: Adjacent angles on a straight line sum to 180°

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Problem 1:

We have a right angle (90°) split into two parts: one is 70°, and the other is $ y^\circ $.
Since they form a right angle:
$$
y + 70^\circ = 90^\circ
\Rightarrow y = 90^\circ - 70^\circ = 20^\circ
$$

✔ Answer: $ y = 20^\circ $

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Problem 2:

Two intersecting lines form vertical angles. One angle is $ 170^\circ $, and the opposite angle is $ t^\circ $.
Vertical angles are equal:
$$
t = 170^\circ
$$

✔ Answer: $ t = 170^\circ $

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Problem 3:

A straight line with two angles: $ b^\circ $ and $ 68^\circ $. They form a linear pair (add to 180°):
$$
b + 68^\circ = 180^\circ
\Rightarrow b = 180^\circ - 68^\circ = 112^\circ
$$

✔ Answer: $ b = 112^\circ $

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Problem 4:

Three angles at a point: $ x^\circ $, $ 35^\circ $, and $ 35^\circ $. All three add up to 180° (since they're on a straight line or forming a triangle-like structure). But wait — let's look carefully.

Actually, it looks like these three angles form a straight line (linear pair), so their total is 180°:
$$
x + 35^\circ + 35^\circ = 180^\circ
\Rightarrow x + 70^\circ = 180^\circ
\Rightarrow x = 110^\circ
$$

✔ Answer: $ x = 110^\circ $

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Problem 5:

Two angles: $ 110^\circ $ and $ 2k^\circ $, and they appear to be supplementary (on a straight line).
$$
2k + 110^\circ = 180^\circ
\Rightarrow 2k = 70^\circ
\Rightarrow k = 35^\circ
$$

✔ Answer: $ k = 35^\circ $

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Problem 6:

Two intersecting lines. One angle is labeled $ (d + 70)^\circ $, and the adjacent angle is $ 130^\circ $. These two form a linear pair (they’re adjacent and on a straight line), so:
$$
(d + 70) + 130 = 180
\Rightarrow d + 200 = 180
\Rightarrow d = 180 - 200 = -20
$$

Wait! That gives $ d = -20 $, which is impossible for an angle.

Let’s double-check.

Actually, if $ (d + 70)^\circ $ and $ 130^\circ $ are vertical angles, then they are equal:
$$
d + 70 = 130
\Rightarrow d = 60
$$

But in the diagram, the $ (d+70)^\circ $ is shown next to $ 130^\circ $, and they are adjacent — likely forming a straight line. So they should be supplementary.

So:
$$
(d + 70) + 130 = 180
\Rightarrow d + 200 = 180
\Rightarrow d = -20
$$

This can't be right. So perhaps I misread.

Wait — maybe $ (d + 70)^\circ $ and $ 130^\circ $ are not adjacent. Let’s consider that $ 130^\circ $ is one angle, and $ (d + 70)^\circ $ is the vertical angle to it? Then they are equal.

But in standard notation, vertical angles are across from each other.

Looking at typical layout: when two lines cross, opposite angles are equal.

So if one angle is $ 130^\circ $, its vertical angle is also $ 130^\circ $. If $ (d + 70)^\circ $ is that vertical angle, then:
$$
d + 70 = 130 \Rightarrow d = 60
$$

Yes — this makes sense.

Alternatively, if $ (d + 70)^\circ $ is adjacent to $ 130^\circ $, they would add to 180°, but that leads to negative $ d $, which is invalid.

So likely, $ (d + 70)^\circ $ is the vertical angle to $ 130^\circ $, meaning:
$$
d + 70 = 130 \Rightarrow d = 60
$$

✔ Answer: $ d = 60^\circ $

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✔ Final Answers:

1. $ y = 20^\circ $
2. $ t = 170^\circ $
3. $ b = 112^\circ $
4. $ x = 110^\circ $
5. $ k = 35^\circ $
6. $ d = 60^\circ $

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