In trapezium PQRS, the sides PQ and SR are parallel. Angle RSP is 120 . and RS = SP = 1/3PQ. What is the size of angle PQR ?

I made a sketch, and let RS = PS = x

then QP = 3x
Draw RC || SP to meet QP at C
Then CP = x and QC = 2x
PCRS is now a ||gram, making angle QCR = 60°
In triangle RCQ, by the cosine law:
QR^2 = (2x)^2 + x^2 - 2(x)(2x)cos60°
= 5x^2 - 4x^2 (1/2)
= 3x^2
QR = √3 x

In the same triangle, by the sine law:
sinQ/x = sin60/(√3 x)
sinQ = xsin60/(√3x)
= sin60/√3
= (√3/2)/√3 = 1/2
Q = 30°

Well, it seems like PQRS got itself into a bit of a geometry circus act! Let's see if we can figure out the size of angle PQR.

Since PQ and SR are parallel, we know that angle PQR and angle RSP are corresponding angles. Corresponding angles are congruent when the lines are parallel. So, angle PQR must be 120 degrees as well!

It looks like PQRS is doing some synchronized acrobatics with equal angles. Now that's what I call a balanced trapezium!

Since PQ and SR are parallel, we can conclude that ∠PQR is equal to ∠SRP. This is due to the property that opposite angles formed by a transversal crossing parallel lines are equal.

Given that ∠RSP = 12°, we can determine the value of ∠SRP.

Since RS = SP, it implies that triangle RSP is isosceles. In an isosceles triangle, the angles opposite the equal sides are equal.

Therefore, ∠SRP = ∠SPR = 180° - 120° = 60°.

Now, we know that ∠SRP = ∠PQR = 60°.

So, the size of angle PQR is 60°.

To find the size of angle PQR in trapezium PQRS, we need to use the information given in the problem.

First, we know that sides PQ and SR are parallel, so we can conclude that angle PQR is equal to angle RSP. This is because when a transversal (RP) intersects two parallel lines (PQ and SR), the corresponding angles are equal.

Next, we are given that angle RSP is 120 degrees. Since angle PQR is equal to angle RSP, we can conclude that angle PQR is also 120 degrees.

Therefore, the size of angle PQR is 120 degrees.