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Quantitative Aptitude

Triangles

In Δ ABC , D ...

Question

In $Δ$ ABC, D,E and F are taken on AB, AC and BC respectively, so that EF=CF and DF=BF. If $∠$ A=40 $^{\circ}$ , then find $∠$ DFE?

100^o

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120^o

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90^o

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140^o

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Solution

The correct option is A 100^o

Option (a)

Take DBF = a $^{\circ}$ and ECF = b $^{\circ}$ .
40 + a $^{\circ}$ + b $^{\circ}$ = 180 $\Rightarrow$ a $^{\circ}$ + b $^{\circ}$ = 140
DFB + EFC + DFE = 180 $\Rightarrow$ 180 – 2a $^{\circ}$ + 180 – 2b $^{\circ}$ + DFE = 180 $\Rightarrow$ 180 $^{\circ}$ – 2(a + b) = - DFE
DFE = 280 – 180 = 100 $^{\circ}$

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Question 46. In triangle DEF shown below, points A, B, and C are taken on DE, DF and EF respectively such that EC = AC and CF = BC. If angle D = 40 degrees, then what is angle ACB in degrees?

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In Δ ABC, D,E and F are taken on AB, AC and BC respectively, so that EF=CF and DF=BF. If ∠A=40∘, then find ∠DFE?

The correct option is A 100o Option (a) Take DBF = a∘ and ECF = b∘. 40 + a∘ + b∘ = 180 ⇒ a∘ + b∘= 140 DFB + EFC + DFE = 180 ⇒ 180 – 2a∘ + 180 – 2b∘ + DFE = 180 ⇒ 180∘ – 2(a + b) = - DFE DFE = 280 – 180 = 100∘

In $Δ$ ABC, D,E and F are taken on AB, AC and BC respectively, so that EF=CF and DF=BF. If $∠$ A=40 $^{\circ}$ , then find $∠$ DFE?