Question

In the given triangle ACB is the right angle triangle at C and CD is perpendicular to AB if < B= 40° ,the find
1) <BAC
2) <ACD
3) < BCD

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Answer 1

Given: <C=90°,

<B=40°

To find : 1)<BAC

2)<ACD

3)<BCD

Solutions : 1)<BAC

<BAC+<ACB+<CBA=180°

<BAC+90°+40°=180°

<BAC=180°-130°

<BAC=50°

2)<ACD

<ACD+<CDA+<DAC=180°

<ACD+90°+50°=180°

<ACD=180°-140°

<ACD=40°

3)<BCD

<BCD+<CDB+<DBC=180°

<BCD+90°+40°=180°

<BCD=180°-130°

<BCD=50°