Angle Addition Postulate Worksheets - Math Monks - Free Printable

Let's solve each problem on the Angle Addition Postulate Worksheet step by step.

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🔷 Angle Addition Postulate Recap:

> If point $ D $ lies in the interior of $ \angle ABC $, then:
> $$
> m\angle ABC = m\angle ABD + m\angle DBC
> $$
> This means that the measure of a larger angle is the sum of the measures of two adjacent smaller angles that make it up.

We'll apply this postulate to each problem.

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Problem 1:

> If $ m\angle ABD = 56^\circ $ and $ m\angle DBC = 60^\circ $, find $ m\angle ABC $

From the diagram:
- Point $ D $ is between $ A $ and $ C $, so $ \angle ABC $ is made up of $ \angle ABD $ and $ \angle DBC $.

$$
m\angle ABC = m\angle ABD + m\angle DBC = 56^\circ + 60^\circ = \boxed{116^\circ}
$$

✔ Answer: $ \boxed{116^\circ} $

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Problem 2:

> If $ m\angle MLN = 15^\circ $ and $ m\angle NLO = 28^\circ $, find $ m\angle MLO $

From the diagram:
- $ \angle MLO $ is composed of $ \angle MLN $ and $ \angle NLO $
- So:

$$
m\angle MLO = m\angle MLN + m\angle NLO = 15^\circ + 28^\circ = \boxed{43^\circ}
$$

✔ Answer: $ \boxed{43^\circ} $

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Problem 3:

> If $ m\angle ABD = 90^\circ $, $ m\angle ABC = 5x - 6 $, and $ m\angle CBD = 3x $, find $ m\angle ABC $

From the diagram:
- $ \angle ABD $ is made up of $ \angle ABC $ and $ \angle CBD $
- So:
$$
m\angle ABD = m\angle ABC + m\angle CBD
$$
Substitute values:
$$
90 = (5x - 6) + 3x
$$
$$
90 = 8x - 6
$$
Add 6 to both sides:
$$
96 = 8x
$$
Divide by 8:
$$
x = 12
$$

Now plug back into $ m\angle ABC = 5x - 6 $:
$$
5(12) - 6 = 60 - 6 = \boxed{54^\circ}
$$

✔ Answer: $ \boxed{54^\circ} $

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Problem 4:

> If $ m\angle MLO = 91^\circ $ and $ m\angle NLO = 20^\circ $, find $ m\angle MLN $

From the diagram:
- $ \angle MLO $ is made up of $ \angle MLN $ and $ \angle NLO $
- So:
$$
m\angle MLO = m\angle MLN + m\angle NLO
$$
$$
91^\circ = m\angle MLN + 20^\circ
$$
$$
m\angle MLN = 91^\circ - 20^\circ = \boxed{71^\circ}
$$

✔ Answer: $ \boxed{71^\circ} $

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Problem 5:

> If $ m\angle OPQ = 158^\circ $, $ m\angle OPR = 9x + 16 $, and $ m\angle RPQ = 7x - 2 $, find $ m\angle OPR $

From the diagram:
- $ \angle OPQ $ is made up of $ \angle OPR $ and $ \angle RPQ $
- So:
$$
m\angle OPQ = m\angle OPR + m\angle RPQ
$$
$$
158 = (9x + 16) + (7x - 2)
$$
Simplify:
$$
158 = 16x + 14
$$
Subtract 14:
$$
144 = 16x
$$
Divide by 16:
$$
x = 9
$$

Now find $ m\angle OPR = 9x + 16 $:
$$
9(9) + 16 = 81 + 16 = \boxed{97^\circ}
$$

✔ Answer: $ \boxed{97^\circ} $

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Problem 6:

> If $ m\angle FCD = x + 41 $, $ m\angle BCF = x + 78 $, and $ m\angle BCD = 95^\circ $, find $ x $

From the diagram:
- $ \angle BCD $ is made up of $ \angle BCF $ and $ \angle FCD $
- So:
$$
m\angle BCD = m\angle BCF + m\angle FCD
$$
$$
95 = (x + 78) + (x + 41)
$$
$$
95 = 2x + 119
$$
Subtract 119:
$$
-24 = 2x
$$
$$
x = -12
$$

Wait — negative angle? Let’s double-check.

But wait! The total angle $ \angle BCD = 95^\circ $, but $ \angle BCF = x + 78 $, and $ \angle FCD = x + 41 $. Adding them gives $ 2x + 119 $. But if $ x = -12 $, then:
- $ \angle BCF = -12 + 78 = 66^\circ $
- $ \angle FCD = -12 + 41 = 29^\circ $
- Sum: $ 66 + 29 = 95^\circ $ → ✔ Correct!

Even though $ x $ is negative, the angle measures are valid. So it's acceptable.

✔ Answer: $ \boxed{-12} $

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✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | $ \boxed{116^\circ} $ |
| 2 | $ \boxed{43^\circ} $ |
| 3 | $ \boxed{54^\circ} $ |
| 4 | $ \boxed{71^\circ} $ |
| 5 | $ \boxed{97^\circ} $ |
| 6 | $ \boxed{-12} $ |

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