in the isosceles trapezium angle c=115, find the remaining angles

Hi!

 

Let us consider the following isosceles trapezium ABCD

Whose AD = BC, AB||CD and ∠DCB = 115°. To find the measure remaining angles of the trapezium, you may extend CD to E such that DE = AB

 

In trapezium ABCD, DC||AB

∴ ∠ABC + ∠DCB = 180°

⇒∠ABC = 180° – 115° = 65°

It is given that AD = BC  …(1)

According to the construction done, DE = AB and DE || AB

Thus, ABED is a parallelogram

∴ AD = BE  … (2)

From (1) and (2)

BC = BE

⇒∠BEC = ∠BCE = 180° – ∠BCD = 180° – 115° = 65°

⇒∠BED = 65°

ABED is a parallelogram

∴∠A = ∠BED = 65°

And ∠D = 180° – ∠BED = 180° – 65° = 115°

 

Hope! You got the answer.

Cheers!