Question

I'll mark best answer as brainliest!​

Answer 1

Answer:

Hope it helps you! Refer the attachment too!

Step-by-step explanation:

Sol :- Given ABCD is a rhombus and EDC

is an equal Δ and ∠DAB=48∘

→EDC is equal Δ so 

∠E=∠C=∠D=60∘

→∠A=∠C=48∘ (in rhombus)

→∠A+∠D=180∘

⇒48∘+∠D=180∘

⇒∠D=132=∠B

Now in ΔECB which is isosceles Δ

∠C=60+48=108∘

and ∠E=∠B=2180−∠C

⇒∠E=∠B=2180−108 

⇒∠E=∠B=36

Hence ∠BEC=36∘ --Ans (i)

Now for (ii)∠DEB=∠DEC−∠BEC

=60−36

∠DEB=24∘

for (iii) ∠BFC in ΔBFC

∠BFC=180∘−∠ECB−∠FBC

∠BFC=