Question

please solve this problem.....​

Answer 1

Answer:

40° & 65°

Step-by-step explanation:

In the given figure,

∠BAC = 75° ; CE ║ AB ; ∠ECD = 40°

we have to find the other 2 angles of the ΔABC

let ∠ECA = x

if i consider AC as the transversal through the parallel lines AB and CE, then we have the interior alternate angles equal,

ie, ∠BAC = ∠ECA

=> 75° = x

now we have ∠BCA,∠ECA & ∠ECD , 3 angles formed on the straight line, BD,

=> ∠BCA+∠ECA +∠ECD = 180°

=> ∠BCA + 75°+40° = 180°

=> ∠BCA + 115° = 180°

=> ∠BCA = 180° - 115° = 65°

=> ∠BCA = 65°

now in ΔABC, ∠ABC ,  ∠BAC & ∠BCA angles are formed,

we know ∠ABC + ∠BAC + ∠BCA = 180°

=> ∠ABC+75°+65° = 180°

=> ∠ABC+140° = 180°

=> ∠ABC = 180° - 140°

=> ∠ABC = 40°

Answer 2

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