# 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°.

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

**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°.

**☛ Related Questions:**

- In Fig. 6.39, sides QP and RQ of ∆PQR are produced to points S and T respectively. If ∠SPR =135° and ∠PQT = 110°, find ∠PRQ.
- In Fig. 6.40, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ∠XYZ, find ∠OZY and ∠YOZ.
- In Fig. 6.42, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.
- In Fig. 6.43, if PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65° then find the values of x and y.

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