Question

Can someone pls tell me how to do this

Answer 1

Answer:

<x=135°

<Y=45°

Step-by-step explanation:

∆EFB is an isosceles triangle as EF=EB

Hence, <EFB=<EBF=let them be equal to z

<E=<F=<A=<T=90°

[all angles in a rectangle are 90°]

in∆EFB

<E+<F+<B=180°

[angle sum property of triangles]

90°+z+z=180°

2z=90°

z=45°

<EFB=<EBF=45°

Now,

<EBF+<X=180°

45°+<x=180°

<x=135°

<F=<EFB+<Y

90°=45°+<Y

<Y=45°

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