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In a triangle \(ABC , ∠ BCA =50°\). D and E are points on AB and AC, respectively, such that AD = DE. If F is a point on BC such that BD = DF, then \(∠ FDE\), in degrees, is equal to
In a triangle \(ABC , ∠ BCA =50°\). D and E are points on AB and AC, respectively, such that AD = DE. If F is a point on BC such that BD = DF, then \(∠ FDE\), in degrees, is equal to
Updated On: Aug 27, 2024
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The Correct Option is B
Approach Solution - 1
We know that, angle on straight line equal to 180°.
so,\( 180-2a+x+180-2b=180\)
\(x-2a-2b+180=0\)
\(x=2a+2b-180…..(1)\)
In \(△ABC, a+b+50=180\) [ ∴ sum of the angles of triangle=180°]
\(= a+b=130\)
Put the value of a+b in equation (1)
\(x= 2(a-b)-180\)
\(x= 2(130)-180\)
\(x= 260-180\)
\(x=80°\)
∴ The △FDE, in degree, is equal to 80°.
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Approach Solution -2
From the triangle ABC,
\(∠A + ∠B + ∠C = 180\degree\)
\(\angle A + \angle B + 50\degree = 180\degree\)
\(∠A + ∠B = 130\degree\)
In the quadrilateral CFDE,
\(∠C + ∠F + ∠D + ∠E = \)3600
500 + 1800 \(- ∠A + ∠x +\) 1800 \(- ∠B =\) 3600
500 \(+ ∠x = ∠A + ∠B\)
500 \(+ ∠x =\) 1300
\(∠x =\) 800
\(∠FDE =\) 800
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