The diagram shows triangle PQR. PSR is a straight line.
size of angle PQS : size of angle SQR = 3:1
S
Work out the size of angle PSQ

Angle PSQ is 61°

Step-by-step explanation:

In the triangle ΔPQR, we have;

m∠QPS = 65°

m∠PRQ = 43°

Line PSR is a straight line

m∠PQS : m∠SQR = 3:1

Therefore;

∠PQR = 180° - (∠QPS + ∠PRQ), by the angle sum theorem of a triangle

∴ ∠PQR = 180° - (65° + 43°) = 72°

∠PQR = 72°

∠PQR = m∠PQS + m∠SQR by angle addition postulate

m∠PQS : m∠SQR = 3:1

∴ m∠PQS = 3 × m∠SQR

∴ ∠PQR = m∠PQS + m∠SQR = 3 × m∠SQR + m∠SQR = 4 × m∠SQR

∠PQR = 4 × m∠SQR

∴ 4 × m∠SQR = ∠PQR = 72°

m∠SQR = 72°/4 = 18°

m∠PQS = 3 × m∠SQR = 3 × 18° = 54°

m∠PQS = 54°

∠PSQ = 180° - (∠QPS + ∠PQS) by the angle sum theorem

∴ ∠PSQ = 180° - (65° + 54°) = 61°

Angle ∠PSQ = 61°.

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.