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# In Fig. 6.32, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y

**Solution:**

Given: AB || CD, ∠APQ = 50° and ∠PRD = 127°

To Find: x and y

When a ray intersects a line, the sum of adjacent angles formed is 180°.

When two parallel lines are cut by a transversal, alternate interior angles formed are equal.

AB and CD are parallel lines cut by transversal PQ hence the alternate interior angles formed are equal.

∠APQ = ∠PQR and hence x = 50°.

Similarly, AB and CD are parallel lines cut by transversal PR hence the alternate angles formed are equal.

∠APR = ∠PRD = 127°

∠APQ + ∠QPR = ∠PRD = 127°

50° + y = 127°

y = 127° - 50°

y = 77°

**☛ Check: **NCERT Solutions Class 9 Maths Chapter 6

**Video Solution:**

## In Fig. 6.32, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y

NCERT Maths Solutions Class 9 Chapter 6 Exercise 6.2 Question 5

**Summary:**

In Fig. 6.32, if AB || CD, ∠APQ = 50°, and ∠PRD = 127°, then the values of x and y are 50° and 77° respectively.

**☛ Related Questions:**

- In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.
- In Fig. 6.30, if AB||CD, EF ⊥ CD and ∠GED =126°, find ∠AGE, ∠GEF and ∠FGE.
- In Fig. 6.31, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.[Hint: Draw a line parallel to ST through point R.]
- In Fig. 6.33, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB||CD.

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