Question

solve the given question
#anayuso

$AB = 16CM DE=15CM \sin( \alpha = \frac{3}{5} ) \sin( \beta = \frac{4}{5} ) \\ find \: length \: of \: bd$
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Answer 1

Answer:

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Step-by-step explanation:

$To \: find = length \: of \: BD \\ Given = \\ AB = 16.cm \\ DE = 15.cm \\ \alpha = \frac{3}{5} \\ \\ \beta = \frac{4}{5} \\ \\ so \: firstly \: using \\ tan \: ACB = \frac{AC}{CB} \\ \\ ie \: tan \: \beta = \frac{AC}{CB} \\ \\ \frac{4}{5} = \frac{16}{CB} \\ \\ \frac{1}{5} = \frac{4}{CB} \\ \\ CB = 5 \times 4 \\ ie \: CB = 20.cm$

$now \: using \\ tan \: ECD = \frac{ED}{CD} \\ \\ ie \: tan \: \alpha = \frac{CD}{CD} \\ \\ \frac{3}{5} = \frac{15}{CD} \\ \\ \frac{1}{5} = \frac{5}{CD} \\ \\ CD = 5 \times 5 \\ ie \: CD = 25.cm$

$now \: thus \: then \\ BD=BC+CD \: \: \: \: \: \: \: \: \: \: ( \: B - C - D \: ) \\ = 20 + 25 \\ = 45.cm \\ \\ hence \: BD=45.cm$