In Fig. 5.16, PA || BC || DT and AB || DC. Then, the values of a and b are respectively
a. 60°, 120°
b. 50°,130°
c. 70°,110°
d. 80°,100°

Solution:

Given, PA || BC || DT and AB || DC

We need to find the values of a and b.

Since, PA and BC are parallel lines cut by a transversal AB,

We know that the alternate interior angles are equal.

So, ∠ PAB = ∠ ABC

∠a = 50°

Since, AB and DC are parallel lines cut by a transversal BC,

We know that the consecutive interior angles are equal.

So, ∠ ABC + ∠ DCB = 180°

∠a + ∠b = 180°

50° + ∠b = 180°

∠b = 180° - 50°

∠ b = 130°

Therefore, ∠a = 50° and ∠b = 130°

✦ Try This: From the figure, find the value of ∠AOD.

☛ Also Check: NCERT Solutions for Class 7 Maths Chapter 5

NCERT Exemplar Class 7 Maths Chapter 5 Problem 17

In Fig. 5.16, PA || BC || DT and AB || DC. Then, the values of a and b are respectively: a. 60°, 120°, b. 50°,130°, c. 70°,110°, d. 80°,100°

Summary:

In Fig. 5.16, PA || BC || DT and AB || DC. Then, the values of a and b are 50° and 130° respectively

☛ Related Questions:

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