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# Isosceles Triangle Problem

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 $$\angle TSQ$$?

Note by Ian Mana
4 years, 3 months ago

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Let ∠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 · 4 years, 3 months ago

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 · 4 years, 3 months ago

In your fifth line I get lost, you say x + 2 * angle TSQ angle PSQ. What does it mean? · 4 years, 2 months ago

∠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. · 4 years, 2 months ago