Solve fast In Fig 9 42, side BC of △ABC is produced to point D such that - Maths - Triangles

Question

Solve fast

I n   F i g   9 42 ,   s i d e   B C   o f
  △ A B C   i s   p r o d u c e d   t o
  p o i n t   D   s u c h   t h a t   b i
s e c t s   a t   ∠ A B C   a n d   ∠ A
C D   m e e t   a t   p o i n t   E   I f
  ∠ B A C = 68 ° ,   f i n d   ∠ B E
C

Rishabh Mittal · Accepted Answer

Dear Student ,

Please find below the solution to the asked query :

∠BAC=68°In △ABC,∠BAC+∠ABC+∠BCA=180°       Angle Sum Property 68° + ∠ABC+∠BCA=180°∠ABC+∠BCA=112°   
 i∠BCA+∠ACD=180°                     Linear Pair ∠ACD=180°-∠BCADividing both sides by 2, we get∠ACD2 = 180°-∠BCA2∠ACE=180°-∠BCA2      
 iiIn △BCE,∠EBC + ∠BCE  + ∠BEC=180°         Angle Sum Property ∠ABC2 + ∠BCA + ∠ACE + ∠BEC=180°          Since ∠EBC=∠ABC2 , As BE is angle bisector ∠ABC2 + ∠BCA + 180°-∠BCA2 + ∠BEC=180°            Using ii ∠ABC+2 ∠BCA +180°-∠BCA 2 + ∠BEC=180° ∠ABC+ ∠BCA +180°2 + ∠BEC=180°112°+180°2 + ∠BEC=180°               Using i 146° + ∠BEC=180°  ∠BEC=34°       ANS
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Solve fast I n F i g . 9 . 42 , s i d e B C o f △ A B C i s p r o d u c e d t o p o i n t D s u c h t h a t b i s e c t s a t ∠ A B C a n d ∠ A C D m e e t a t p o i n t E . I f ∠ B A C = 68 ° , f i n d ∠ B E C .

Solve fast

$I n F i g . 9.42, s i d e B C o f △ A B C i s p r o d u c e d t o p o i n t D s u c h t h a t b i s e c t s a t ∠ A B C a n d ∠ A C D m e e t a t p o i n t E . I f ∠ B A C = 68 °, f i n d ∠ B E C .$