Textbook - Triangles Questions in English

(N/A) Let $A$ be the tip of the fishing rod and $B$ be the point on the water surface directly below $A$. Let $C$ be the initial position of the fly. Given $AB = 1.8 \, m$ and $BC = 2.4 \, m$.
In the right-angled triangle $\triangle ABC$,by Pythagoras theorem:
$AC^2 = AB^2 + BC^2$
$AC^2 = (1.8)^2 + (2.4)^2 = 3.24 + 5.76 = 9.00$
$AC = \sqrt{9} = 3 \, m$.
So,the length of the string out is $3 \, m$.
She pulls the string at a rate of $5 \, cm/s = 0.05 \, m/s$.
In $12 \, seconds$,the length of the string pulled in is $12 \times 0.05 = 0.6 \, m$.
Let the new position of the fly be $D$ after $12 \, seconds$. The new length of the string is $AD = AC - 0.6 = 3 - 0.6 = 2.4 \, m$.
In the right-angled triangle $\triangle ADB$:
$AD^2 = AB^2 + BD^2$
$(2.4)^2 = (1.8)^2 + BD^2$
$5.76 = 3.24 + BD^2$
$BD^2 = 5.76 - 3.24 = 2.52$
$BD = \sqrt{2.52} \approx 1.587 \, m$.
The total horizontal distance of the fly from Nazima is $BD + 1.2 \, m = 1.587 + 1.2 = 2.787 \, m \approx 2.79 \, m$.