Question

parallelogram ABCD, BE is perpendicular to CD and BF is perpendicular to AD. Given that <EBC=20° FIND <BCD,<DAB,<FBE ​

Answer 1

Step-by-step explanation:

578 is the correct answer

Answer 2

Answer:

The angle $<BCD,<DAB,<FBE$ are $70^0,70^0,140^0$ respectively.

Step-by-step explanation:

Given

parallelogram ABCD, BE is perpendicular to CD and BF is perpendicular to AD.

Angle EBC is $20^0$

From triangle EBC

Angle CEB is $90^0$

Sum of the angle is $180^0$

We get as

$<ECB+20+90=180\\\\<ECB=70^0$

Since Angle C is $70^0$

We know that the adjacent angles of a triangle make $180^0$

So we get as

$<B+70=180\\\\<B=110^0$

And angle A will be $70^0$

From triangle FAB we write as

$<A+<ABF+<BFA=180\\\\70+<ABF+90=180\\\\<ABF=20^0$

The angle B is cut into 3 parts so we write as

$<EBC+<FBE+<ABF=180\\\\20+<FBE+20=180\\\\<FBE=140^0$