Question

In the adjacent figure AB// CD. Find the value of x, y and z

please guys answer it.​

Answer 1

$\bold\SOLUTION$

Given that AB∣∣CD

Taking PQ as transversal,

∠RQP=∠QPC=80

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Now,

∠QPC+2x+3x=180

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80

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+5x=180

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5x=130

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x=26

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now, applying angle sum property in triangle PQR,

2x+y+∠PQR=180

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2×26+80

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+y=180

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y=48

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now,

z=2x+80

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(An exterior angle of a triangle is equal to the sum of the opposite interior angles)

z=2×26

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+80

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=52

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+80

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=132

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z=132

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Answer 2

Answer:

x=20

y=60

z=120

Step-by-step explanation:

Since AB// CD, <DPQ = <PQA= 2x+3x= 5x____________________1

(they are alternate interior angles)

And, <PQA and <PQR is a linear pair.

>> 80+<PQA=180

>>  <PQA= 100

hence, from 1, 5x=100

>> x=100/5= 20

<DPR=<PRQ  (they are alternate interior angles)

>> y= 3x= 3*20 =60

and, y and z are a linear pair.

>> 60+z=180

=z=180-60

>>z=120

Hope I was able to help!!