Question

Please help me solve this

answer is 70

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Answer 1

Answer:

70°

Step-by-step explanation:

• $In\:\triangle ABC,\:\measuredangle A =40\degree$ (Given)

• $Let \:\measuredangle ABC= x \: and \:\measuredangle ACB = y$

• $40\degree + x + y = 180\degree$ (sum of the angles of a triangle)

• $\rightarrow x + y = 180\degree-40\degree$

• $\rightarrow\bold{\red{ x + y = 140\degree}}$....(1)

• $\measuredangle CBD + x = 180\degree$ (angles in linear pair)

• $\rightarrow x = 180\degree-\measuredangle CBD$....(2)

• $\measuredangle BCE + y = 180\degree$ (angles in linear pair)

• $\rightarrow y = 180\degree-\measuredangle BCE$....(3)

• $\rightarrow x+y = 180\degree-\measuredangle CBD+180\degree-\measuredangle BCE$ (Adding equations 2 and 3)

• $\rightarrow 140\degree= 360\degree-(\measuredangle CBD+\measuredangle BCE)$

• $\rightarrow\measuredangle CBD+\measuredangle BCE= 360\degree- 140\degree$

• $\rightarrow\measuredangle CBD+\measuredangle BCE= 220\degree$

• $\rightarrow\frac{1}{2}(\measuredangle CBD+\measuredangle BCE)= \frac{1}{2}*220\degree$ (Dividing throughout by 2)

• $\rightarrow\frac{1}{2}\measuredangle CBD+\frac{1}{2}\measuredangle BCE= 110\degree$

• $\rightarrow \measuredangle CBO+\measuredangle BCO= 110\degree$....(4) (BO and CO are bisectors of angle CBD and angle BCE respectively)

• $In\:\triangle OBC$

• $\measuredangle CBO+\measuredangle BCO+\measuredangle BOC= 180\degree$ (Sum of the angles of a triangle)

• $\rightarrow 110\degree+\measuredangle BOC= 180\degree$ (From equation 4)

• $\rightarrow \measuredangle BOC= 180\degree-110\degree$

• $\rightarrow \huge{\orange{\measuredangle BOC= 70\degree}}$