In triangle ΔABC, ∠A+∠ABC+∠BCA=180o 34o+∠ABC+30o=180o ∠ABC=116o So, ∠ABC+∠CBD=180o 116o+∠CBD=180o ∠CBD=180o−116o ∠CBD=64o Now use remote interior angle theorem. So, ∠X=∠CBD+∠D=64o+45o=109o
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