Where is the oil?

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By writing cosine law for $\Delta ABC$ we get :

$\quad\displaystyle l_{1}^2=l_{2}^2+d^2-2dl_{2}\cos{(\frac{\pi}{2}-\alpha)}$

Now, since $l_1,l_2\gg d$ we can write this equation as:

$\quad l_{1}^2=l_{2}^2-2dl_{2}\cos{(\frac{\pi}{2}-\alpha)}$

From this equation we can find $\sin{\alpha}$ directly as:

$\displaystyle\quad\sin{\alpha}=\frac{l_{2}^2-l_{1}^2}{2dl_{2}}\quad\quad\small{\color{#20A900}{\cos{(\frac{\pi}{2}-\alpha)}=\cos{\frac{\pi}{2}}\cos{\alpha}+\sin{\frac{\pi}{2}}\sin{\alpha}=\sin{\alpha}}}$

By writing $l_{2}^2-l_{1}^2$ as $(l_2-l_1)(l_1+l_2)$ , we get:

$\displaystyle\quad\sin{\alpha}=\frac{(l_2-l_1)(l_1+l_2)}{2dl_{2}}=\frac{(l_1-l_2)}{d}\frac{(l_1+l_2)}{2l_2}$

Again because $l_1,l_2\gg d$ we can say that $\frac{l_1}{l_2}\approx 1$ , because of this $\frac{(l_1+l_2)}{2l_2}\approx1$ .

And finaly $l_2-l_1=c\tau$ , by adding this we get:

$\displaystyle\quad\sin{\alpha}=\frac{c\tau}{d}\quad\quad\quad \Rightarrow \quad\quad\quad \boxed{\alpha=\arcsin{\frac{c\tau}{d}}}$

D=20cm so D=0.2m and the time delay physically means $y=Dsin\alpha$ and the delay $\tau =0.2 ns$ so $y= c\tau$ so $\alpha = asin(\frac{y}{D})$ yields $\alpha = asin(\frac{c\tau}{D})$ yields $\alpha = 17.457$