Question

In the given alongside figure, AD=AB=AC, BD is parallel to CA and angle ACB = 65° . Find angle DAC.
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Answer 1

Answer:

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Answer 2

Answer:

∠DAC=130°

Step-by-step explanation:

BY THE PARALLEL SIDES

∠ACB+∠DBC=180°

65°+∠DBC=180°

∠DBC=180°-65°=115°

∠ABD +65°=115°

∠ABD=115°-65°=50

BY ISOCELES TRIANGLE PROPERTY IN ΔABD:

∠ADB=∠ABD=50°

IN ΔABD-

∠ADB+∠ABD+∠BAD=180°

50°+50°+∠BAD=180°

∠BAD=180°-100°=80°  ............[1]

NOW IN ΔABC-

∠ABC+∠BAC+∠ACB=180°

∠BAC+65°+65°=180°

∠BAC=180°-130

∠BAC=50°  .......[2]

BY ADDING BOTH-

∠BAC+∠BAD=50°+80°

∠DAC=130°

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