Ray AB∥ ray CE, ∠APC = 100°, ∠BAD= 56°, Find m∠PCD = ?
Answers
Given:
Ray AB∥ ray CE
∠APC = 100°
∠BAD= 56°
To find:
m∠PCD
Solution:
Since AB // CE and AD is a transversal line
∴ ∠BAD = ∠PDC = 56° ........ [alternate angles] ...... (i)
From the figure, we get
∠APC + ∠CPD = 180° ....... [Linear Pair]
substituting ∠APC = 100° (given)
⇒ 100° + ∠CPD = 180°
⇒ ∠CPD = 180° - 100°
⇒ ∠CPD = 80° ...... (ii)
Now, consider Δ PDC, we get
∠PCD + ∠PDC + ∠CPD = 180° ...... [angle sum property]
substituting from (i) & (ii), we get
⇒ ∠PCD + 56° + 80° = 180°
⇒ ∠PCD + 136° = 180°
⇒ ∠PCD = 180° - 136°
⇒ ∠PCD = 44°
Thus, m∠PCD = 44°.
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