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Question
In a regular pentagon PQRST, PR = QT intersect at N. Find the angle RQT and QNP.
Solution
Each interior angle of a regular pentagon
= `((5 - 2) xx 180°)/(5)`
= 3 x 36°
= 108°
Now, In ΔPQT,
⇒ PT = PQ ....(sides of a regular pentagon)
∠PQT = ∠PTQ = x(say)
Now, ∠QT + ∠{TQ +∠QPT = 180°
⇒ x + x + 108° = 180°
⇒ 2x = 72°
⇒ x = 36°
⇒ ∠PQT = ∠PTQ = 36°
Similarly, we can prove that in ΔPQR,
∠QPR = ∠QRP = 36°
Now, ∠RQT
= ∠RQP -∠PQT
= 108° - 36°
= 72°
In ΔQNP,
∠PQN +∠QPN + ∠QNP = 180°
⇒ 36° + 36° + ∠QNP = 180°
⇒ ∠QNP = 180° - 72°
⇒ ∠ QNP = 108°.
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