pLS ANSWER QUESTION 2
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Answer:
( B) 70, is answer. I think it will help you
Answered by
1
Answer:
In △DEF
∠EDF+∠DEF+∠DFE=180
o
⇒ ∠DEF+∠DFE=180
o
−∠EDF
⇒ ∠DEF+∠DFE=180
o
−40
o
=140
o
......(i)
In △BCF, CB=CF
⇒ ∠CFB=∠CBF (Isos.△CBF)
or ∠EFD=∠CBF.......(ii)
In △AEC,CA=CE⇒∠CEA=∠CAE (Isos. △ prop.)......(iii)
or ∠DEF=∠CAE
∴ from (i),(ii),(iii)
∠CAE+∠CBF=140
o
180
o
−∠CAD+180
o
−∠CBD=140
o
⇒ ∠CAD+∠CBD=360
o
−140
o
=220
o
In quad, DACB
∠ADB+∠CAD+∠CBD+∠ACB=360
o
⇒ 40
o
+220
o
+∠ACB=360
o
⇒ ∠ACB=360
o
−260
o
=100
o
Explanation:
hope it helps u...
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