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Question

In figure, $ \angle \mathrm{BAC}=90°$, $ \mathrm{AD}\perp \mathrm{BC}$ and $ \angle \mathrm{BAD}=50°$, then $ \angle \mathrm{ACD}$ is


O

A

50°

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B

40°

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C

70°

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D

60°

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Solution

The correct option is A

50°


The explanation for the correct answer.

Find the required angle:

Given: BAC=90° and BAD=50°

Therefore,

DAC=BAC-BADDAC=90°-50°DAC=40°

Now in triangle ADC,

ACD+DAC+ADC=180° (Sum angle property of triangle)

ACD+40°+90°=180ACD+130°=180°ACD=180°-130°ACD=50°

Hence option A is correct.


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