In a right angled $\triangle ABC$, $\angle B = 90^\circ$, $\angle C = 15^\circ$ and $|AC| = 7.\;$ Let a point $D$ (Random Point) be taken on $AC$ and then perpendicular lines $DE$ and $DF$ are drawn on $AB$ and $AC$ respectively. What is the probability of $DE\cdot DF >3?$
Attempt:
By trigonometry, I got the length of other two side from the hypotenuse $AC:$
$AB$ $\approx 1.8117$
$BC$ $\approx 6.7614$. And than, I got the equation that
$DE\cdot DF = (6.7614 – DE)\cdot AE\;$ (from the similarity of both $\triangle AED$ and $\triangle DFC$)
Again, from right angled $\triangle AED$,
$\dfrac{AE}{DE} = \tan 15^{\circ}\quad \implies \quad AE = DE\cdot \tan 15^\circ$
Here, I got stuck. I couldn't find a way out to proceed and skip that situation. I became lost and was unable to complete that process. Any kind of help or clue will be greatly helpful for me to step forward.
Best Answer
We have :- $$ o = x \sin 15 \tag{1}$$ $$a= y \ cos 15 \tag{2} $$ Multiplying $(1),(2)$ , we get :- $$ o\cdot a = xy \sin 15 \cos 15 = \frac{xy}{2} sin 30 = \frac{xy}{4} > 3 $$ ( $\because 2 \sin \theta \cos \theta = \sin 2\theta $)
Hence , we need $xy = x(7-x) > 12$ or $x^2-7x+12<0$
As this is of the form $ax^2+bx+c$ , and since $a>0$ , the expression is negative between the roots . Therefore, we must have $x \in (3,4) $
$\therefore $ The probability $\frac{4-3}{7} = \frac{1}{7}$