Question

Find the length of the shortest straight beam that will reach to the side of the building from the ground outside the wall.

Non-right triangles and trigonometry
ANSWERED
asked 2021-09-09
The 5 feet wall stands 28 ft froom the building. Find the length of the shortest straight beam that will reach to the side of the building from the ground outside the wall.

Expert Answers (1)

2021-09-10
From similar triangles,
\(\displaystyle{x}={6}\)
\(\displaystyle{\left({28}+{x}\right)}={h}\)
\(\displaystyle{H}=\frac{{{28}+{x}}}{{x}}\times{6}\)
Now, from the \(\displaystyle\triangle{A}{B}{C}\),
\(\displaystyle{A}{C}^{{2}}+{B}{C}^{{2}}={A}{B}^{{2}}\)
\(\displaystyle{L}^{{2}}={\left({x}+{28}\right)}^{{2}}+{\left(\frac{{{28}+{x}}}{{x}}\times{6}\right)}^{{2}}\)
\(\displaystyle\frac{{{d}{L}}}{{\left.{d}{x}\right.}}={0}\)
\(\displaystyle{2}{L}\times\frac{{{d}{L}}}{{\left.{d}{x}\right.}}=\frac{{d}}{{\left.{d}{x}\right.}}{\left({x}^{{2}}+{28}^{{2}}+{56}{x}+{\left(\frac{{{x}^{{2}}+{28}^{{2}}+{56}}}{{x}^{{2}}}\times{36}\right)}\right)}\)
\(\displaystyle{2}{L}\times\frac{{{d}{L}}}{{\left.{d}{x}\right.}}={0}={2}{x}+{0}+{56}+{0}+{\left({28}^{{2}}\times{36}{x}\times\frac{{-{2}}}{{x}^{{3}}}\right)}+{56}\times{36}{x}{\left(-\frac{{1}}{{x}^{{2}}}\right)}\)
\(\displaystyle{2}{x}+{56}{448}{\left(-\frac{{1}}{{x}^{{3}}}\right)}–\frac{{2016}}{{x}^{{3}}}={0}\)
\(\displaystyle{2}{x}^{{4}}–{56448}–{2016}{x}={0}\)
\(\displaystyle{X}={14}\times{3762}{\quad\text{or}\quad}{x}=-{11}\times{377}\)
The shortest possible length is \(\displaystyle{14}\times{3762}.\)
0
 
Best answer

expert advice

Have a similar question?
We can deal with it in 3 hours
...