Question

pls tell the value of x from the picture​

Answer 1

Answer:

Hi there!!

It's answer is 55.

Answer 2

$\huge\boxed{\bold{\blue{Solution:-}}}$

Given :-

AB||CD and EF||CD

And $\angle CEF=150^{\circ}$

$\angle BCE=25^{\circ}$

According to question :-

1st we have to find $\sf \angle ECD$

Such that EF|| CD

Then :-

$\sf\angle CEF+\angle ECD=180^{\circ}$(consecutive)

$\sf \angle ECD+150^{\circ}=180^{\circ}\\ \sf\angle ECD=180^{\circ}-150^{\circ}\\ \sf\angle ECD=30^{\circ}$

Produce CD to G such that

GC||AB

Then :-

$\sf\angle GCB+\angle BCE+\angle ECD =180^{\circ}$ (Linear pair)

$\sf \angle </p><p>GCB+25^{\circ}+30^{\circ}=180^{\circ}\\ \sf\angle GCB=180^{\circ}-55^{\circ}\\ \sf\angle GCB= 125^{\circ}$

Now

$\sf \angle GCB+x=180^{\circ}$(consecutive angle )

$\leadsto 125^{\circ}+x=180^{\circ}\\ \sf\rightarrow x=180^{\circ}-125^{\circ}\\ \sf\implies x=55^{\circ}$

You can also find angle by this method

$\sf \angle ABC= \angle BCE + \angle ECD$ (linear pair )

$\leadsto x=25^{\circ}+20^{\circ}\\ \sf\leadsto x=55^{\circ}$

Answer:- (x= 55°)

$\sf {Hope \:it's \:help !}$