Question

In Fig. 8.52, AB = AC; BC = CD, ZBAF = 130° and DE || BC.
Calculate :
(1) ZCDE (ii) ZDCE.
Fig. 8pls answer this step by step pls​

Answer 1

Step-by-step explanation:

(i) angle CDE- 50°

(ii) angle DCE- 15°

see angle FAD+angle EAD=180°

so angle EAD=50°

now angle ACB+ABC+CAB=180°(it's a ∆)so as it is an isosceles triangle angle ACB and angle ABC=65°.

next we know that DE II BC, so angle AED and angle ADE are equal to 65°.

now as ∆BCD is an isosceles triangle. Angle CBD and Angle CDB are equal

to 65°.now the angles----ADE+CDE+CDB=180°

i.e. 65°+CDE+65°=180°

so as per calculation ANGLE CDE=50°.

next AED +CED=180° so angle CED = 115°.now the angles add up to 180°

CDE+CED+DCE=180°

50°+115°+DCE=180°.

so after calculation angle DCE turns out to be 15°.

Please follow me back.i wrote this for 10 minutes continuously.

Answer 2

Answer:

sorry mate I don't know

Step-by-step explanation:

Don't worry