Question

Answer this question
Chapter : Triangles
Class : 9

No Spam​

Answer 1

GIVEN=>

AB=AC, /_A=42° and /_ACD=18°

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FIND=>

/_BCD=?

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SOLUTION=>

in ∆ ADC tatal angle 180°

/_A + /_ACD+/_ADC=180°(∆rules)

42°+18°+/_ADC=180°

/_ADC=180°-60°

/_ADC=120°

/_ADC+/_BDC=180°

/_BDC=180°-/_ADC

/_BDC=180°-120°

/_BDC=60

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AB=AC (GIVEN)

/_ABC=/_ACB--(1)

IN ∆ABC

/_A + /_ACB+/_ABC=180°(∆rules)

42°+ /_ABC+/_ABC=180°(EQ(1)

/_ABC=/_ACB)

2/_ABC=180°-42°

/_ABC=138°/2

/_ABC=69°

/_ABC=/_ACB=69°

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in ∆BCD

/_DBC+/_BCD+/_CDB=180°(∆rules)

/_BCD=180°-/_DBC-/_CDB

/_BCD=180°- 69°-60°

/_BCD=180°-129°

/_BCD=51°

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answer

/_BCD=51°

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I hope it helps plz mark as brainlist

Answer 2

Answer:

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