\(\displaystyle∠{A}≅∠{D}\) since all right triangles are congruent and \(\displaystyle∠{B}{C}{A}≅∠{E}{C}{D}\) because vertical angle pairs are congruent. So, \(\displaystyle△{A}{B}{C}∼△{D}{E}{C}\) by A.

Similar figures have corresponding side lengths that are proportional. If x is the distance across the river (which is BC¯¯¯¯¯), then we can write:

\(\displaystyle{B}\frac{{C}}{{E}}{C}={A}\frac{{C}}{{D}}{C}\)

PSKx/EC=48/0.4

To find EC, use Pythagorean Theorem on \(\displaystyle△{D}{E}{C}\):

\(\displaystyle{D}{E}^{{2}}+{D}{C}^{{2}}={E}{C}^{{2}}\)

\(\displaystyle{0.3}^{{2}}+{0.4}^{{2}}={E}{C}^{{2}}\)

\(\displaystyle{0.25}={E}{C}^{{2}}\)

0.5=EC

Hence,

\(\displaystyle\frac{{x}}{{0.5}}=\frac{{48}}{{0.4}}\)

\(\displaystyle\frac{{x}}{{0.5}}{\left({0.5}\right)}=\frac{{48}}{{0.4}}{\left({0.5}\right)}\)

x=60 mi

Similar figures have corresponding side lengths that are proportional. If x is the distance across the river (which is BC¯¯¯¯¯), then we can write:

\(\displaystyle{B}\frac{{C}}{{E}}{C}={A}\frac{{C}}{{D}}{C}\)

PSKx/EC=48/0.4

To find EC, use Pythagorean Theorem on \(\displaystyle△{D}{E}{C}\):

\(\displaystyle{D}{E}^{{2}}+{D}{C}^{{2}}={E}{C}^{{2}}\)

\(\displaystyle{0.3}^{{2}}+{0.4}^{{2}}={E}{C}^{{2}}\)

\(\displaystyle{0.25}={E}{C}^{{2}}\)

0.5=EC

Hence,

\(\displaystyle\frac{{x}}{{0.5}}=\frac{{48}}{{0.4}}\)

\(\displaystyle\frac{{x}}{{0.5}}{\left({0.5}\right)}=\frac{{48}}{{0.4}}{\left({0.5}\right)}\)

x=60 mi