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Label the rays (lower and left first): a as PQ, b as RS, c as WX and d as YZ. Also mark the points of intersection as A (intersection of PQ & WX), B (intersection of WX & RS), C (intersection of RS & YZ), and D (intersection of PQ & YZ).
`m/_PDC` is 85°, `m/_WAD` is 95°, `m/_ABC` is 72°, `m/_BCD` is y and `m/_ADC` is x.
Since line segment DA stands on ray a,
`m/_DAB`` +` `m/_DAW` = 180° [linear pair]
Consider the rays/lines WX and YZ, cut by the transversal PQ at points A and D respectively. The corresponding angles `/_PDC` and `/_DAB` are equal (both measuring 85°). Hence, according to the parallel postulate, WX ll YZ.
Here, line segment CD stands on ray d,
` `` ` ` ` `m/_PDC+x` = 180° [linear pair]
`rArr` x=`m/_ADC` = 180°-`m/_PDC` = 180°-85°=95°
Again, consider the pair of parallel lines WX and YZ cut by the transversal RS at points B and C respectively. So, the corresponding angles `/_ABC` and `/_DCR` must be equal.
`m/_DCR` is thus 72°.
and line segment DC stands on ray b, hence
y+ `m/_DCR` = 180° [linear pair]
`rArr` y = 180-72 =108°
Therefore, the required solution is: x = 95 and y = 108.
In the figure a is a straight line that intersects the straight lines c and d. The measure of the combined angle formed by x and 85 is equal to 180 degrees. This gives the angle x = 180 - 85 = 95.
For lines c and d with transversal a, the alternate interior angles are equal. This gives c and d as parallel lines. As consecutive interior angles add up to 180, y = 180 - 72 = 108.
The angle x = 95 degrees and y = 108 degrees.
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