Question

please solve guys. pls​

Answer 1

Answer:

Shown that, ∠PAQ = 60°

Step-by-step explanation:

As shown in the attached figure, O is the centre of the given circle.  OR and OS are joined.

From ΔAPQ, ∠PAQ + ∠APQ + ∠AQP = 180°

=> ∠PAQ + ∠PRO + ∠QSO = 180°  [In ΔPRO, OP=OR radius of same circle,

                                                         ∴ ∠APQ=∠PRO, similarly, ∠AQP=∠QSO]

=> ∠PAQ + (180° - ∠ARO) + (180° - ∠ASO) = 180°

=> ∠PAQ + 180° - ∠ARO + 180° - ∠ASO = 180°

=> ∠PAQ + 180° - (∠ARS + ∠ORS) - (∠ASR + ∠OSR) = 0

=> ∠PAQ + 180° - (∠ARS + 60°) - (∠ASR + 60°) = 0  [In ΔORS all sides are equal to the radius of the given circle.  ∴ Angles are all equal and must be 60° each]

=> ∠PAQ + 180° - ∠ARS - 60° - ∠ASR  - 60° = 0

=> ∠PAQ + 60° - (∠ARS + ∠ASR) = 0

=> ∠PAQ + 60° - (180° - ∠RAS) = 0 [From ΔARS, sum of three angles is 180°]

=> ∠PAQ + 60° - 180° + ∠PAQ = 0

=> 2∠PAQ - 120° = 0

=> 2∠PAQ = 120°

∴ ∠PAQ = 60°