Question

please give me answer of these question
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Answer 1

Answer:

1) angle ACD =120 degree

sum of angle cba and cab =120°

Answer 2

Answer:

1) In triangle ABC,

∠ABC + ∠BCA + ∠CAB =180° ( sum of all the angles of a triangle)

=> 45° + ∠BCA + 75° =180°

=> ∠BCA = 180° - 120°

=> ∠BCA = 60°

Now, ∠ACD= 180° - ∠BCA ( linear pair )

=> ∠ACD = 180° -60°

=> ∠ACD = 120°

2) ∠ACB = 180°-∠ACD ( linear pair )

=> y° = 180° - 130°

=> y° = 50°

In triangle ABC,

∠ABC + ∠BCA + ∠CAB =180° ( sum of all the angles of a triangle)

=> 68° + 50° + x° =180°

=> x° = 180° - 118°

=> x° = 62 °

3) ∠ACB = 180°-∠ACD ( linear pair )

=> y° = 180° - 65°

=> y° = 115°

In triangle ABC,

∠ABC + ∠BCA + ∠CAB =180° ( sum of all the angles of a triangle)

=> x° + 115° + 32° =180°

=> x° = 180° - 147°

=> x° = 33°