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/_ACD and /_DCB form a linear pair. In this diagram, m/_ACD=[?]^(@) Enter :

∠ACD \angle A C D and ∠DCB \angle D C B form a linear pair.\newlineIn this diagram, m∠ACD=[?]∘ m \angle A C D=[?]^{\circ} \newlineEnter :

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Full solution

Q. ∠ACD \angle A C D and ∠DCB \angle D C B form a linear pair.\newlineIn this diagram, m∠ACD=[?]∘ m \angle A C D=[?]^{\circ} \newlineEnter :
  1. Linear Pair Angles: Linear pair angles add up to 180180 degrees, so m/∠ACD+m/∠DCB=180∘m/_{\angle ACD} + m/_{\angle DCB} = 180^\circ.
  2. Find Angle ACD: We know m/DCBm/_{DCB}, but we need to find m/ACDm/_{ACD}. Let's say m/ACD=xm/_{ACD} = x and m/DCB=180∘−xm/_{DCB} = 180^\circ - x.
  3. Substitute Values: Now we plug in the value of m/DCBm/_{DCB} into the equation.\newline$x + (\(180\)° - x) = \(180\)°.
  4. Simplify Equation: Simplify the equation: \(x - x + 180° = 180°\).
  5. No Solution: This simplifies to \(180° = 180°\), which doesn't help us find \(x\).

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