Answer 1

Tìm CD?

Có \(\widehat{BAD}\)\(=90^0-15^050'=74^010'\)

\(\widehat{ADB}\)\(=15^050'\)

Có \(\dfrac{BD}{sin\widehat{BAD}}=\dfrac{AB}{sin\widehat{ADB}}\)\(\Leftrightarrow BD=\dfrac{AB.sin\widehat{BAD}}{sin\widehat{ADB}}\)

Lại có \(\widehat{ACB}\)\(=35^040'\);\(\widehat{BAC}\)\(=90^0-35^040'=54^020'\)

Có \(\dfrac{BC}{sin\widehat{BAC}}=\dfrac{AB}{sin\widehat{ACB}}\)\(\Leftrightarrow BC=\dfrac{AB.sin\widehat{BAC}}{sin\widehat{ACB}}\)

\(\Rightarrow CD=BD-BC=\)\(\dfrac{AB.sin\widehat{BAD}}{sin\widehat{ADB}}\)\(-\dfrac{AB.sin\widehat{BAC}}{sin\widehat{ACB}}\)\(=146,5\)(m)

 

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