Question

in figure 7a and 7b, AB||CD. Determine the values of x, y, and Z.​

Answer 1

Answer:

In figure 7a

x=70°

y=30°

z=80°

In figure 7b

x=80°

y=60°

z=40°

I hope it will helpful

Answer 2

The values of x,y, and z in both the figures are as follows-

figure 7a : x = 70°, y= 30° , z = 80°

figure 7b : x = 80°, y=60°, z=40°

Step-by-step explanation:

Given:

line AB || CD

In figure 7a, ∠APQ=70°,∠PRD=100°

In figure 7b, angles on line CD = 80°, 40°

To find:

values of x,y, and z in both the figures

Solution:

For both the figures,

line AB || CD

In figure 7a

x = ∠APQ ..............(angles in alternate pair)

∴ x = 70°

Now, ∠PRD+∠PRQ=180° .......(angles in linear pair)

∴ 100°+∠PRQ=180°

∴ ∠PRQ=180°-100°

∴ ∠PRQ=80°...(1)

In ΔPQR,

∠QPR+∠PQR+∠PRQ= 180° ......(sum of angles of a triangle)

y + x +80° = 180°.........(from 1)

∴ y + 70° +80° = 180°

∴ y + 150° = 180°

∴ y  = 180°- 150°

∴ y  = 30°

Now, ∠APQ+∠QPR+∠BPR= 180°

The angles are in linear pair

∴ 70°+y+z=180°

∴ 70°+ 30°+z=180°

∴ 100°+z=180°

∴ z=180°-100°

∴ z=80°

Thus, for figure 7a the values of x, y, & z are 70°, 30°, and 80° respectively

In figure 7b,

Let's assume the value y as ∠P, and measure 80° & 40° as ∠Q & ∠R respectively

In ΔPQR

∠QPR + ∠PQR + ∠PRQ = 180°

y + 80°+40° = 180°..........(given)

∴ y + 120° = 180°

∴ y = 180°-120°

∴ y = 60°.........(2)

Again, ∠BPD = ∠PDC ........angles in alternate pair

∴ z = ∠PDC = 40°

∴ z = 40°.........(3)

The angles ∠QPA, ∠QPR, & ∠BPD with measures x, y, & z will have a sum of 180° because they are angles in linear pair

∠QPA+∠QPR +∠BPD = 180°

x + y + z = 180°

∴ x + 60°+ 40° = 180°............(from 2 & 3)

∴ x+ 100° = 180°

∴ x =  180°- 100°

∴ x =  80°

Thus, in figure 7b, x = 80°, y = 60°, & z = 40°

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