Question

Circle T has diameters RP and QS. The measure of ∠RTQ is 12° less than the measure of ∠RTS.

Circle T is shown. Line segments T S, T R, T Q, and T P are radii. Lines are drawn to connect points S and R and points P and Q to form secants. Angles R T S and Q T P are congruent.

What is the measure of Arc Q P?

Answer 1

If we consider QS to be the diameter,

If we consider QS to be the diameter,then ∠QTS = 180°

If we consider QS to be the diameter,then ∠QTS = 180°Now, let ∠RTQ = x

If we consider QS to be the diameter,then ∠QTS = 180°Now, let ∠RTQ = xAccording to the question,

If we consider QS to be the diameter,then ∠QTS = 180°Now, let ∠RTQ = xAccording to the question,∠RTQ = ∠RTS - 12°

If we consider QS to be the diameter,then ∠QTS = 180°Now, let ∠RTQ = xAccording to the question,∠RTQ = ∠RTS - 12°⇒ ∠RTS = x + 12°

If we consider QS to be the diameter,then ∠QTS = 180°Now, let ∠RTQ = xAccording to the question,∠RTQ = ∠RTS - 12°⇒ ∠RTS = x + 12°∴ ∠QTS = ∠RTQ + ∠RTS

If we consider QS to be the diameter,then ∠QTS = 180°Now, let ∠RTQ = xAccording to the question,∠RTQ = ∠RTS - 12°⇒ ∠RTS = x + 12°∴ ∠QTS = ∠RTQ + ∠RTS= x + x + 12° = 2x + 12° = 180°

If we consider QS to be the diameter,then ∠QTS = 180°Now, let ∠RTQ = xAccording to the question,∠RTQ = ∠RTS - 12°⇒ ∠RTS = x + 12°∴ ∠QTS = ∠RTQ + ∠RTS= x + x + 12° = 2x + 12° = 180°⇒ 2x = 168°

If we consider QS to be the diameter,then ∠QTS = 180°Now, let ∠RTQ = xAccording to the question,∠RTQ = ∠RTS - 12°⇒ ∠RTS = x + 12°∴ ∠QTS = ∠RTQ + ∠RTS= x + x + 12° = 2x + 12° = 180°⇒ 2x = 168°⇒ x = 84°

If we consider QS to be the diameter,then ∠QTS = 180°Now, let ∠RTQ = xAccording to the question,∠RTQ = ∠RTS - 12°⇒ ∠RTS = x + 12°∴ ∠QTS = ∠RTQ + ∠RTS= x + x + 12° = 2x + 12° = 180°⇒ 2x = 168°⇒ x = 84°⇒ ∠RTQ = 84°

If we consider QS to be the diameter,then ∠QTS = 180°Now, let ∠RTQ = xAccording to the question,∠RTQ = ∠RTS - 12°⇒ ∠RTS = x + 12°∴ ∠QTS = ∠RTQ + ∠RTS= x + x + 12° = 2x + 12° = 180°⇒ 2x = 168°⇒ x = 84°⇒ ∠RTQ = 84°Now,

If we consider QS to be the diameter,then ∠QTS = 180°Now, let ∠RTQ = xAccording to the question,∠RTQ = ∠RTS - 12°⇒ ∠RTS = x + 12°∴ ∠QTS = ∠RTQ + ∠RTS= x + x + 12° = 2x + 12° = 180°⇒ 2x = 168°⇒ x = 84°⇒ ∠RTQ = 84°Now,∵∠QTP and ∠RTS are vertical angles

If we consider QS to be the diameter,then ∠QTS = 180°Now, let ∠RTQ = xAccording to the question,∠RTQ = ∠RTS - 12°⇒ ∠RTS = x + 12°∴ ∠QTS = ∠RTQ + ∠RTS= x + x + 12° = 2x + 12° = 180°⇒ 2x = 168°⇒ x = 84°⇒ ∠RTQ = 84°Now,∵∠QTP and ∠RTS are vertical angles∴ ∠QTP = 84° + 12° = 96°

If we consider QS to be the diameter,then ∠QTS = 180°Now, let ∠RTQ = xAccording to the question,∠RTQ = ∠RTS - 12°⇒ ∠RTS = x + 12°∴ ∠QTS = ∠RTQ + ∠RTS= x + x + 12° = 2x + 12° = 180°⇒ 2x = 168°⇒ x = 84°⇒ ∠RTQ = 84°Now,∵∠QTP and ∠RTS are vertical angles∴ ∠QTP = 84° + 12° = 96°As ∠QTP is the central angle, hence the measure of arc QP is 96°

If we consider QS to be the diameter,then ∠QTS = 180°Now, let ∠RTQ = xAccording to the question,∠RTQ = ∠RTS - 12°⇒ ∠RTS = x + 12°∴ ∠QTS = ∠RTQ + ∠RTS= x + x + 12° = 2x + 12° = 180°⇒ 2x = 168°⇒ x = 84°⇒ ∠RTQ = 84°Now,∵∠QTP and ∠RTS are vertical angles∴ ∠QTP = 84° + 12° = 96°As ∠QTP is the central angle, hence the measure of arc QP is 96°Ans) Measure of arc QP = 96°

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