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Re: In parallelogram PQRS, TR bisects ∠QRS. [#permalink]
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AE wrote:

Please explain ∠QRT + ∠RTQ + ∠QRS= which triangle?


Please explain ∠QRT + ∠RTQ + ∠TQR= triangle TQR.. There was a typo fixed it!
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Re: In parallelogram PQRS, TR bisects ∠QRS. [#permalink]
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as pqrs is parallelogram pq|| sr ---- 1
thus angle(p) = angle(r)
thus r = a
thus angle(qrt) = 1/2*angle(r) = a/2
thus angle(qtr) = angle(trs)
b = a/2
2b = a therefore ans is c
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Re: In parallelogram PQRS, TR bisects ∠QRS. [#permalink]
sandy wrote:
AE wrote:

Please explain ∠QRT + ∠RTQ + ∠QRS= which triangle?


Please explain ∠QRT + ∠RTQ + ∠TQR= triangle TQR.. There was a typo fixed it!

Oh! thanks. I was confused.
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Re: In parallelogram PQRS, TR bisects ∠QRS. [#permalink]
babanya wrote:
as pqrs is parallelogram pq|| sr ---- 1
thus angle(p) = angle(r)
thus r = a
thus angle(qrt) = 1/2*angle(r) = a/2
thus angle(qtr) = angle(trs)
b = a/2
2b = a therefore ans is c


Great idea.
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Re: In parallelogram PQRS, TR bisects ∠QRS. [#permalink]
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Re: In parallelogram PQRS, TR bisects ∠QRS. [#permalink]
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