Question

In fig 6.41 ,if AB II DE ,angle BAC = 35° and angle CDE = 53° ,find angle DCE.​

Answer 1

Step-by-step explanation:

In the given figure ∠BAC = ∠CED [Alternate angles] ⇒ ∠CED = 35° In ∆CDE, ∠CDE + ∠DCE + ∠CED = 180° [Angle sum property of a triangle] ⇒ 53° + ∠DCE + 35° = 180° ⇒ ∠DCE + 88° = 180° ⇒ ∠DCE = 180° – 88° = 92° Hence, ∠DCE = 92° Ans

Answer 2

Step-by-step explanation:

LBAC=LCED ....alternate angles

Therefore ,LCED=35°

LCED+LCDE+LDCE=180 ... sum property of triangle

53°+35°+LDCE=180

LDCE=180-88

LDCE= 92

hope this will help you