In ΔABC,BD bisects angle B. If ∠A=23∠B and ∠B=3∠C, then ∠BDC is:
A
75o
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B
105o
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C
90o
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D
120o
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Solution
The correct option is B105o ∠A=23∠Bor∠B=32∠A Again, ∠A=23∠B=23×3∠Cor∠C=∠A2 ∠A+∠B+∠C=180o Substituting the values of ∠Band∠C ∠A+32∠A+∠A2=180o or∠A=60o ∴∠B=32∠A=32×60o=90o ∠BDC=∠A+12∠B=60o+12×90o=105o