Shortest length of ladder over wall (Find angle)

[grapz]

A wall of height 8 m stands parallel to and 27 m from a tall building . A ladder with its foot on the ground is to pass over the wall and lean on the building. What angle will the shortest such ladder make with the ground?

I tried making a diagram . But i am not sure how to express this and solve so that i achieve a max angle.

thx

 

[morson]

Are you wanting a maximum angle or shortest ladder length? I'll assume the latter:

Let t = the distance from the base of the building to the foot of the ladder.
Let h = the vertical distance of the ladder above the ground.
Let x = the angle the ladder makes with the ground in radians (as differentiation becomes a lot simpler when working with radians).

The distance, a, of the ladder is given by:

\(\displaystyle a = sqrt{t^2 + h^2}\)

By basic trigonometry, \(\displaystyle tan(x) = \frac{h}{t}\\)

Also, by similar triangles, \(\displaystyle \frac{8}{t-27}\ = \frac{h}{t}\\)

Hence, \(\displaystyle tan(x) = \frac{8}{t-27}\\), \(\displaystyle t = 8cotx + 27\)

Similarly: \(\displaystyle h = 8 + 27tanx\)

So substituting these in, we obtain:

\(\displaystyle a = sqrt{64cot^2x + 432cotx + 793 + 432tanx + 729*tan^2x}\)

Now find da/dt, set it equal to 0, and solve for the angle x (the solution should be in radians and the angle should be acute).

 

[galactus]

\(\displaystyle \L\\L=L_{1}+L_{2}=8csc({\theta})+27sec({\theta})\)

\(\displaystyle \L\\\frac{dL}{d{\theta}}=-8csc({\theta})cot({\theta})+27sec({\theta}tan({\theta}), \;\ 0<{\theta}<\frac{\pi}{2}\)

=\(\displaystyle \L\\\frac{-8cos({\theta})}{sin^{2}({\theta})}+\frac{27sin({\theta})}{cos^{2}({\theta})}=\frac{-8cos^{3}({\theta})+27sin^{3}({\theta})}{sin^{2}({\theta})cos^{2}({\theta})}\)

\(\displaystyle \L\\\frac{dL}{d{\theta}}=0 \;\ if \;\ 27sin^{3}({\theta})=8cos^{3}({\theta}), \;\ tan^{3}({\theta})=\frac{8}{27}\)

Now, finish?.