Question

Please answer me this question?

Answer 1

Solution:

Given: DF is tangents to the circle with center O.
∠CAE=62°,∠ACE=43°

To find, value of angle a,b&c<span class="_wysihtml5-temp-placeholder"></span>

In ΔACE,
∠BAE=∠CAE
∠CAE+∠ECA+∠AEC=180°  (Angle sum propanderty of triangles)
          62°+43°+∠AEC=180°
                  105°+∠AEC=180°
                              ∠AEC=180°-105°
                              ∠AEC=75°
∴∠CEF=180°-∠AEC            (sum of the angles in a straight line equals 180°)
   ∠CEF=180°-75°
   ∠CEF=105°

In ΔDEF,
∠DEF+∠EDF+∠DFE=180°
as, ∠DEF=∠CEF,∠DFE=∠BFA

                   105°+b+c=180°
                              b+c=180°-105°
                              b+c=75°....(i)

∠BDC=∠EDF  (vertically opposite angles are equal)
∠BCD=180°- ∠ABF  (sum of the angles in a straight line equals 180°)
∠BCD=180°- a

In ΔBCD,
∠CBD+∠BCD+∠BDC=180°
            43°+(180°-a)+c=180°
                        223°-a+c=180°
                                   c-a=180°-223°
                                   c-a=-43°
                                   a-c=43°..(ii)

In ΔBCD,
EDF=In ΔBAF ,
∠BAF+∠ABF+∠BFA=180°,
                       62°+a+b=180°
                                a+b=180°-62°
                                a+b=118°..(iii)


    by solving the equations,we can get the value of a,b,c.
we get