Question

In given figure AB ||DE LABC =75° and LCDE =145° then find = LBCD​

Answer 1

Answer:

Extend the line AB to F

So, AF || DE

DC is traversal

∠EDC=∠BOD (Alternate interior angles)

∠EDC=∠BOD = 145°

∠BOD+∠BOC =180° (Linear pair)

145°+∠BOC =180°

∠BOC =180°-145°

∠BOC =35°

In BOC

∠BOC+∠BCO = ∠ABC (exterior angle property of triangle)

35°+∠BCO = 75°

∠BCO = 75°-35°

∠BCO =40°

So, ∠BCD =40°

Answer 2

Step-by-step explanation:

Answer:

Answer:∠BCD =40°

Answer:∠BCD =40°Step-by-step explanation:

Answer:∠BCD =40°Step-by-step explanation:Refer the attached figure

Answer:∠BCD =40°Step-by-step explanation:Refer the attached figureExtend the line AB to F

Answer:∠BCD =40°Step-by-step explanation:Refer the attached figureExtend the line AB to FSo, AF || DE

Answer:∠BCD =40°Step-by-step explanation:Refer the attached figureExtend the line AB to FSo, AF || DEDC is traversal

Answer:∠BCD =40°Step-by-step explanation:Refer the attached figureExtend the line AB to FSo, AF || DEDC is traversal∠EDC=∠BOD (Alternate interior angles)

Answer:∠BCD =40°Step-by-step explanation:Refer the attached figureExtend the line AB to FSo, AF || DEDC is traversal∠EDC=∠BOD (Alternate interior angles)∠EDC=∠BOD = 145°

Answer:∠BCD =40°Step-by-step explanation:Refer the attached figureExtend the line AB to FSo, AF || DEDC is traversal∠EDC=∠BOD (Alternate interior angles)∠EDC=∠BOD = 145°∠BOD+∠BOC =180° (Linear pair)

Answer:∠BCD =40°Step-by-step explanation:Refer the attached figureExtend the line AB to FSo, AF || DEDC is traversal∠EDC=∠BOD (Alternate interior angles)∠EDC=∠BOD = 145°∠BOD+∠BOC =180° (Linear pair)145°+∠BOC =180°

Answer:∠BCD =40°Step-by-step explanation:Refer the attached figureExtend the line AB to FSo, AF || DEDC is traversal∠EDC=∠BOD (Alternate interior angles)∠EDC=∠BOD = 145°∠BOD+∠BOC =180° (Linear pair)145°+∠BOC =180°∠BOC =180°-145°

Answer:∠BCD =40°Step-by-step explanation:Refer the attached figureExtend the line AB to FSo, AF || DEDC is traversal∠EDC=∠BOD (Alternate interior angles)∠EDC=∠BOD = 145°∠BOD+∠BOC =180° (Linear pair)145°+∠BOC =180°∠BOC =180°-145°∠BOC =35°

Answer:∠BCD =40°Step-by-step explanation:Refer the attached figureExtend the line AB to FSo, AF || DEDC is traversal∠EDC=∠BOD (Alternate interior angles)∠EDC=∠BOD = 145°∠BOD+∠BOC =180° (Linear pair)145°+∠BOC =180°∠BOC =180°-145°∠BOC =35°In BOC

Answer:∠BCD =40°Step-by-step explanation:Refer the attached figureExtend the line AB to FSo, AF || DEDC is traversal∠EDC=∠BOD (Alternate interior angles)∠EDC=∠BOD = 145°∠BOD+∠BOC =180° (Linear pair)145°+∠BOC =180°∠BOC =180°-145°∠BOC =35°In BOC∠BOC+∠BCO = ∠ABC (exterior angle property of triangle)

Answer:∠BCD =40°Step-by-step explanation:Refer the attached figureExtend the line AB to FSo, AF || DEDC is traversal∠EDC=∠BOD (Alternate interior angles)∠EDC=∠BOD = 145°∠BOD+∠BOC =180° (Linear pair)145°+∠BOC =180°∠BOC =180°-145°∠BOC =35°In BOC∠BOC+∠BCO = ∠ABC (exterior angle property of triangle)35°+∠BCO = 75°

Answer:∠BCD =40°Step-by-step explanation:Refer the attached figureExtend the line AB to FSo, AF || DEDC is traversal∠EDC=∠BOD (Alternate interior angles)∠EDC=∠BOD = 145°∠BOD+∠BOC =180° (Linear pair)145°+∠BOC =180°∠BOC =180°-145°∠BOC =35°In BOC∠BOC+∠BCO = ∠ABC (exterior angle property of triangle)35°+∠BCO = 75°∠BCO = 75°-35°

Answer:∠BCD =40°Step-by-step explanation:Refer the attached figureExtend the line AB to FSo, AF || DEDC is traversal∠EDC=∠BOD (Alternate interior angles)∠EDC=∠BOD = 145°∠BOD+∠BOC =180° (Linear pair)145°+∠BOC =180°∠BOC =180°-145°∠BOC =35°In BOC∠BOC+∠BCO = ∠ABC (exterior angle property of triangle)35°+∠BCO = 75°∠BCO = 75°-35°∠BCO =40°

Answer:∠BCD =40°Step-by-step explanation:Refer the attached figureExtend the line AB to FSo, AF || DEDC is traversal∠EDC=∠BOD (Alternate interior angles)∠EDC=∠BOD = 145°∠BOD+∠BOC =180° (Linear pair)145°+∠BOC =180°∠BOC =180°-145°∠BOC =35°In BOC∠BOC+∠BCO = ∠ABC (exterior angle property of triangle)35°+∠BCO = 75°∠BCO = 75°-35°∠BCO =40°So, ∠BCD =40°

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