Question 1155496

use {{{cos(60)}}} and {{{u=8sqrt(3)}}} to find hypothenuse {{{s}}}

{{{cos(60)=8sqrt(3)/s}}}

{{{s=8sqrt(3)/cos(60)}}}

{{{s=8sqrt(3)/(1/2)}}}

{{{s=2*8sqrt(3)}}}

{{{s=16sqrt(3)}}}

then, using Pythagorean theorem we have

{{{t^2=s^2-u^2}}}

{{{t^2=(16sqrt(3))^2-(8sqrt(3))^2}}}

{{{t^2=256*3-64*3}}}

{{{t^2=192*3}}}

{{{t^2=576}}}

{{{t=sqrt(576)}}}

{{{t=24}}}