# In Fig. 6.41, if AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.

**Solution:**

Given: AB || DE, ∠BAC = 35° and ∠CDE = 53°

To find: ∠DCE

We know that when two parallel lines are cut by a transversal, alternate interior angles formed are equal.

According to angle sum property of a triangle, sum of the interior angles of a triangle is 180°.

Since, AB || DE and AE is the transversal,

∠DEC = ∠BAC [Alternate interior angles]

Thus, ∠DEC = 35°

Now, in △CDE

∠CDE + ∠DEC + ∠DCE = 180° [Angle sum property of a triangle]

53° + 35° + ∠DCE = 180°

∠DCE = 180° - 88°

Thus, we have ∠DCE = 92°.

**Video Solution:**

## In Fig. 6.41, if AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.

### NCERT Maths Solutions Class 9 - Chapter 6 Exercise 6.3 Question 3:

**Summary:**

In Fig. 6.41, if AB || DE, ∠BAC = 35°, and ∠CDE = 53°, then the value of ∠DCE from the given figure is 92°.