Question

answer fast and correctly!!​

Answer 1

Answer:

a + b + c - d + e - f = 144°

Step-by-step explanation:

Given that, ∠SPQ = 90°

PR is a diameter, so ∠RSP = 90°

∴ PQRS is rectangle.  And a + f = 180° - 90° = 90°

∴ c = e  [alternate interior angles]

In ΔOQR, OQ = OR, ∴ e = $\frac{1}{2}$(180° - 108°) = 36°

b = e   [Angle intercepted by Arc PQ on A and R, both on the circumference]

∴ b = 36°

∠QOR = 2∠QAR  [Angle intercepted by Arc QR on the centre is twice that of on any point on the circumference (point A in this case)]

∴ ∠QAR = $\frac{1}{2}$ . 108° = 54°

∴ ∠QPR = ∠QAR = 54°  [Angle intercepted by Arc QR on A and P, both on the circumference]

∠POQ = 180° - 108° = 72°

In ΔOBQ, ∠OBQ = 180° - ∠PBQ = 180° - 90° = 90°

And ∠QOR = ∠BQO + ∠OBQ

∴ d = 108° - 90° = 18°

In ΔOAQ, OA = OQ  [radius of same circle]

∠OAQ = ∠OQA = d = 18°

∴ ∠AOP = 90° - 18° = 72°

∴ ∠ARP = $\frac{1}{2}$ . ∠AOP = $\frac{1}{2}$ . 72° = 36°

a = c + ∠ARP  [since, a is exterior angle of ΔPRD ]

∴ a = 36° + 36° = 72°

∴ f = 90° - 72° = 18°

∴ a + b + c - d + e - f = 72° + 36° + 36° - 18° + 36° - 18° = 144°