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smithjohn

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In ΔPQR, ∠PQR = 70° and S and T are points on PQ and PR respectively,

so that ∠RQT = 55° and ∠QRS = 40°. Show that ∠PST is 35°
 

jimmysmith560

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Is there any similarity involved in this question?

The only angle I'm struggling to prove is ∠TSR.
 

smithjohn

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Is there any similarity involved in this question?

The only angle I'm struggling to prove is ∠TSR.
No all they give us is that info. I got all angles, worked out angle PSR = 110 and angle PTQ = 125 but how can you find the angle PST. I have no clue all the answer says is 35 degrees
 

jimmysmith560

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I'm thinking finding ∠TSR or ∠STQ is most likely what we need to do next. I don't see what else can be done.
 

smithjohn

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I'm thinking finding ∠TSR or ∠STQ is most likely what we need to do next. I don't see what else can be done.
I found angle PSR = 110 and PTQ = 125 using exterior angle of a triangle but still doesnt help
 

jimmysmith560

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RSQ is also isosceles I believe.

If we can find the value of any one of these: <PTS, <TSR, <STQ, that should do it.
 
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jimmysmith560

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I think I found it:

- ΔTRQ is isosceles (∠QTR = ∠RQT = 55°)
Therefore QR = TR

- ΔSRQ is isosceles (∠QSR = ∠SQR = 70°)
Therefore QR = SR

But QR = TR
Therefore TR = SR, meaning ΔSRT is isosceles

- ∠TRS = 30°
Therefore ∠RTS = ∠RST = 75°

Now that we have the value of ∠RST, we can show that ∠PST is 35°

∠QSR = 70°
∠RTS = 75°

- PQ is a straight line:
Therefore: 180 - (70+75) = 35

Therefore ∠PST is 35°
 
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