Question

IfAB II DE, ∠BAC = 35° and ∠CDE= 53°, find ∠DCE. In Fig. 6.41.

Answer 1

hey here is your answer____

we can write that angle BAE=AED=35°

so in triangle CDE

angle C+ angleD+angle E=180

angleDCE=180-53-35=92°

Answer 2

Hello mate ☺

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Solution:

It is given that AB∥DE. Therefore, ∠BAC=∠DEC              (Alternate Interior Angles)

∠BAC=35°      (Given)

Therefore, ∠DEC=35°

In ∆DCE, we have

∠DEC+∠CDE+∠DCE=180°         (Sum of three angles of a triangle =180°)

⇒35°+53°+∠DCE=180°

⇒∠DCE=180°−35°−53°=92°

I hope, this will help you.☺

Thank you______❤

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