I want to know the best answer of the question from Lines and Angles chapter of class 9th ncert math. The question from exercise 6.3of math. Give me the easy way for solving this question of 3 In Fig. 6.41, if AB DE, BAC = 35° and CDE = 53°, find DCE.
We know that AE is a transversal since AB DE
Here BAC and AED are alternate interior angles.
Hence, BAC = AED
It is given that BAC = 35°
AED = 35°
Now consider the triangle CDE. We know that the sum of the interior angles of a triangle is 180°.
∴ DCE+CED+CDE = 180°
Putting the values, we get
DCE+35°+53° = 180°
Hence, DCE = 92°