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Let ∠A be '∅' (see figure.)


\frac{Opposite}{Adjacent} = \frac{3}{4}AdjacentOpposite=43
So, hypothenuse will be 5 by Pythagoras theorem.
Now,

\frac{ 4 * \frac{3}{5} - \frac{4}{5} + 1 }{ 4 * \frac{3}{5} + \frac{4}{5} - 1 }4∗53+54−14∗53−54+1
\frac{ \frac{12- 4 + 5}{5} }{ \frac{12 +4 - 5}{5}}512+4−5512−4+5
5 will cancel out,
= \frac{13}{11}=1113
= 1.18=1.18


\frac{Opposite}{Adjacent} = \frac{3}{4}AdjacentOpposite=43
So, hypothenuse will be 5 by Pythagoras theorem.
Now,

\frac{ 4 * \frac{3}{5} - \frac{4}{5} + 1 }{ 4 * \frac{3}{5} + \frac{4}{5} - 1 }4∗53+54−14∗53−54+1
\frac{ \frac{12- 4 + 5}{5} }{ \frac{12 +4 - 5}{5}}512+4−5512−4+5
5 will cancel out,
= \frac{13}{11}=1113
= 1.18=1.18
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