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### Ladder, Cube & Wall Problem

A ladder of length $l=2\sqrt 6$ unit is resting against a wall just touching the cube of edge 1 unit as shown in the figure. Find h above the cube.

We have, $\tan \theta = h$ and $\sin \theta = \frac{{h + 1}}{l}$

So, $2\sqrt 6 \sin \theta = 1 + \tan \theta$

$\Rightarrow \sqrt 6 (2\sin \theta \cos \theta ) = \cos \theta + \sin \theta$

Squaring, ${(\sqrt 6 \sin 2\theta )^2} = {(\cos \theta + \sin \theta )^2}$

$\Rightarrow 6{\sin ^2}2\theta = 1 + \sin 2\theta$

Let, $\sin 2\theta = t$

We have, $6{t^2} - t - 1 = 0$

$\Rightarrow 6{t^2} - 3t + 2t - 1 = 0$

$\Rightarrow 3t(2t - 1) + (2t - 1) = 0$

$\Rightarrow (3t + 1)(2t - 1) = 0$

Thus, $t = \frac{1}{2} = \sin 2\theta$

[Since, $\sin 2\theta >0$ the other solution is rejected.]

$\therefore 2\theta = {30^ \circ },{150^ \circ }$

Or, $\theta = {15^ \circ },{75^ \circ }$

Now, $h = \tan \theta = \tan {15^ \circ },\tan {75^ \circ } = 2 \pm \sqrt 3$

### A man starts walking from the point P (-3, 4) ....

A man starts walking from the point P (-3, 4), touches the x-axis at R, and then turns to reach at the point Q (0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50 [(PR)^2 + (RQ)^2 ]$ is equal to _ _ _ _ . Solution For time to be minimum at constant speed, the directions must be symmetric. In other words, the angles made by PR and RQ with the vertical must be the same just like in the law of reflection in optics. $tan \theta = \frac {MP}{MR} = \frac {NQ}{NR}$ $\Rightarrow \frac {3-r}{4} = \frac {r}{2}$ $\Rightarrow r=1$ So, $R \equiv ( - 1,0)$ Now, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$