Could triangle jkl be congruent to triangle xyz? explain.

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Ответ:

shondra20

27.04.2023

Mathematics

$\frac{1}{7}$

Step-by-step explanation:

Given

tanx = $\frac{3}{4}$ = $\frac{opposite}{adjacent}$

This is a 3- 4- 5 right triangle with hypotenuse = 5, then

cosx = $\frac{adjacent}{hypotenuse}$ = $\frac{4}{5}$

sinx = $\frac{opposite}{hypotenuse}$ = $\frac{3}{5}$

Then

$\frac{cosx-sinx}{cosx+sinx}$

= $\frac{\frac{4}{5}-\frac{3}{5} }{\frac{4}{5}+\frac{3}{5} }$

= $\frac{\frac{1}{5} }{\frac{7}{5} }$

= $\frac{1}{5}$ × $\frac{5}{7}$

= $\frac{1}{7}$

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