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find ∠OQR ∠RST ∠TRQ
∠ROQ = 48


Find OQR RST TRQ ROQ 48 class=

Sagot :

Answer:

Given

  • ∠ROQ = 48°

Find

  • ∠OQR, ∠RST,  ∠TRQ

Solution

We see that SR is a diameter, therefore Δ SRT is right triangle with ∠STR = 90°

∠ROQ, ∠RST and ∠TRQ all intercept the same arc RT

∠ROQ is twice the ∠RST and ∠TRQ as the central angle:

  • ∠RST = ∠ROQ = 2*∠TRQ ⇒ ∠RST = ∠TRQ = 48°/ 2 = 24°

PQ is tangent to circle, therefore:

  • PQ⊥SR ⇒ ∠OQR = 90°

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