# Question #3a151

May 29, 2017

Use $\sin \theta$ where $\theta$ is the angle of the ladder.
$\Delta B C = 4 \left(\sqrt{3} - \sqrt{2}\right)$

#### Explanation:

Alright. First we draw a diagram. I have labelled the points of the triangle $A$, $B$, and $C$. I will also be referring to the angle $\angle B A C$ as $\theta$ We know that in the first diagram, $\theta = {60}^{\circ}$ and $A B = 8$.

We know that in the second diagram, $\theta = {45}^{\circ}$ and $A B = 8$ as the ladder's length does not change.

Using the trig ratio $\sin \theta = \left(\text{opposite")/("hypotenuse}\right)$

I can say in both diagrams, $\sin \theta = \frac{B C}{A B}$

In the first diagram, $\sin 60 = \frac{B C}{8}$
$B C = 8 \sin 60$

In the second diagram, $\sin 45 = \frac{B C}{8}$
$B C = 8 \sin 45$

Now, we can see that the change in height is just:
$8 \sin 60 - 8 \sin 45 = 8 \cdot \frac{\sqrt{3}}{2} - 8 \cdot \frac{1}{\sqrt{2}}$
$= 4 \sqrt{3} - 4 \sqrt{2}$
$= 4 \left(\sqrt{3} - \sqrt{2}\right)$

$\Delta B C = 4 \left(\sqrt{3} - \sqrt{2}\right)$

As to where the values came from, there's a table for it.
If you are up to a challenge, you can try deriving where the exact values of the trig functions came from. May 29, 2017

$\sin 60 = \frac{h}{8}$.....where h is the height of the top of the ladder from ground.

so, $h = 8 \sin 60$

=> $h = 8 \frac{\sqrt{3}}{2} \implies 4 \sqrt{3}$

When the ladder makes an angle of 45 degrees we have:

$\sin 45 = \frac{h}{8}$

i.e. $h = 8 \sin 45$

=> $h = 8 \frac{\sqrt{2}}{2} \implies 4 \sqrt{2}$

Hence, distance of slip is:

$4 \sqrt{3} - 4 \sqrt{2}$

i.e. $4 \left(\sqrt{3} - \sqrt{2}\right)$

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