In the given figure, if AB = AC, then what is the value of ∠ACB – ∠DBF?

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consider **triangle ABC**

given **angle BAC=40 degree** and **AB=AC** -------(1)

therefore triangle ABC is an isosceles triangle

We know, in an isosceles triangle, the angles opposite to equal sides are equal.

therefore,

**angle ABC=angle ACB** -------(2)

But ,

**angle ABC+angle ACB+angle BAC=180 degree
2( angle ABC)+ 40 degree=180 degree [by (1) and (2)]
2(angle ABC)=180 degree-40 degree=140 degree
angle ABC=140/2=70 degree**

Therefore,

**angle ABC=angle ACB=70 degree ----(3)**

But clearly we can see that, angle CAB and angle DBF are corresponding angles.

therefore ,

**angle ABC= angle DBF=40 [ Reason: corresponding angles are equal ] ----(4)**

By (3) and (4),

**angle ACB- angle DBF= 70 degree- 40 degree= 30 degree**

ANSWER=30 degree

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