A \(\triangle PQR\) is an isosceles triangle where \(PQ=PR\). point \(T\) and \(S\) is in \(PQ\) and \(QR\) respectively, so that \(\triangle PTS\) is an isosceles triangle with \(PT=PS\). if \(\angle RPS\) is \(20^\circ\), Find \(\angle TSQ\)?

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TopNewestLet ∠TPS= x ,∠PTS=PST=z ,∠PQR=∠PRQ=y.

Thus we have,

2y=180-(20+x), or y= \(\frac{180-(20+x)}{2}\).

Again, 2z=(180-x). or, Z= \(\frac{(180-x)}{2}\).

Now in △TQS, we have, ∠STP the external angle. Thus, z= y+ ∠TSQ or, ∠TSQ=\(\frac {(180-x)}{2}\)- \(\frac{180-(20+x)}{2}\) or, ∠TSQ= 10 – Bodhisatwa Nandi · 4 years, 3 months ago

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10 degrees.

let angle PRQ = angle PQR = x

angle PTS = x + angle TSQ

=> angle PST = x + angle TSQ

angle PSQ = angle PST + angle TSQ = x + 2 * angle TSQ angle PSQ = x + 20 (ext angle of triangle PSR)

=> 2 * angle TSQ = 20 => angle TSQ = 10 – Ethan Tan · 4 years, 3 months ago

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– Jordi Bosch · 4 years, 2 months ago

In your fifth line I get lost, you say x + 2 * angle TSQ angle PSQ. What does it mean?Log in to reply

∠PRQ = ∠PQR = a , because isosceles triangle.

∠PTS = ∠PST = b , because isosceles triangle.

∠TSQ = c

∠PSR = 180-(20+a) , because angles in a triangle add up to 180. Or, ∠PSR = 180-(b+c) , because angles in a straight line is 180.

Therefore we get 180-(20+a) = 180-(b+c), or simply, 20+a = b+c. Or in terms of a we get: a = b+c-20

∠QTS = 180-(a+c) . Or, ∠QTS = 180-b . Therefore we get a+c=b . Or in terms of a we get : a = b-c .

We now have two equations, (1) a = b+c-20 and (2) a = b-c

So, b+c-20 = b-c (Subtract b from both sides) c-20 = -c (Add c to both sides) 2c-20 = 0 Therefore, 2c = 20 or simply c = 10

The question asks for ∠TSQ which we named c. We found out c is 10 degrees. So answer is 10. – Mohammad Al Ali · 4 years, 2 months ago

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