Question

Find the measures of the missing sides of 45° - 45° - 90° right triangle with 4cm hypotenuse.

Answer 1

$\sin( \angle{a}) = \frac{bc}{ac}$

$\sin( 45 \degree) = \frac{ bc}{4}$

$\frac{1}{ \sqrt{2} } = \frac{bc}{4}$

$bc = \frac{4}{ \sqrt{2} }$

$bc = 2 \sqrt{2} \: cm$

${ac}^{2} = {ab}^{2} + {bc}^{2}$

${4}^{2} = {ab}^{2} + {2 \sqrt{2} }^{2}$

$16 = {ab}^{2} + 8$

${ab}^{2} = 16 - 8$

${ab}^{2} = 8$

$ab = \sqrt{8}$

$ab = 2 \sqrt{2} \: cm$