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find the measures of ∠adb and ∠bdc formed between the escalators. m∠adb…

Question

find the measures of ∠adb and ∠bdc formed between the escalators.
m∠adb = \boxed{\space}°
m∠bdc = \boxed{\space}°

Explanation:

Step1: Identify supplementary angles

∠ADB and ∠BDC are supplementary? Wait, no, looking at the diagram, ∠ADB and the angle (6x + 10)° are vertical? Wait, no, actually, ∠ADB (4x°) and (6x + 10)° are supplementary? Wait, no, maybe they are adjacent and form a linear pair. Wait, the angles on a straight line sum to 180°. So (6x + 10) + 4x = 180? Wait, no, maybe ∠ADB is 4x and the other angle is (6x + 10), and they are supplementary. Wait, let's check:

Wait, the two angles ∠ADB (4x°) and the angle adjacent to it (6x + 10)° are supplementary (since they form a linear pair, on a straight line). So:

$$6x + 10 + 4x = 180$$

Step2: Solve for x

Combine like terms:

$$10x + 10 = 180$$

Subtract 10 from both sides:

$$10x = 170$$

Divide by 10:

$$x = 17$$

Step3: Find m∠ADB

m∠ADB = 4x° = 4*17 = 68°

Step4: Find m∠BDC

Since ∠ADB and ∠BDC are supplementary (linear pair), m∠BDC = 180° - 68° = 112°? Wait, no, wait, maybe the angle (6x + 10) is equal to ∠BDC? Wait, no, let's re-examine. Wait, maybe the angle (6x + 10) and ∠BDC are vertical angles? Wait, no, the diagram shows that ∠ADB is 4x, and the angle opposite to (6x + 10) is ∠BDC? Wait, no, perhaps I made a mistake. Wait, let's re-express:

Wait, the two angles (6x + 10)° and 4x° are adjacent and form a linear pair, so their sum is 180°. So:

6x + 10 + 4x = 180

10x + 10 = 180

10x = 170

x = 17

Then, ∠ADB is 4x = 68°, and ∠BDC is equal to (6x + 10)° because they are vertical angles? Wait, no, if ∠ADB and (6x + 10) are supplementary, then ∠BDC would be equal to (6x + 10) because ∠ADB and ∠BDC are supplementary? Wait, no, let's check:

If ∠ADB is 4x = 68°, then (6x + 10) = 6*17 + 10 = 102 + 10 = 112°. Then, ∠ADB (68°) and ∠BDC (112°) are supplementary (68 + 112 = 180), which makes sense because they form a linear pair. So yes, that's correct.

Answer:

m∠ADB = $\boxed{68}$°
m∠BDC = $\boxed{112}$°