In the figure, the value of $\angle P$ will be: 
Step 1: Observe the two triangles
The two triangles $\triangle ABC$ and $\triangle PQR$ are given. Their sides are in proportion:
\[
AB : PQ = AC : PR = BC : QR
\]
Check:
\[
\frac{AB}{PQ} = \frac{3.8}{7.6} = \frac{1}{2},
\frac{AC}{PR} = \frac{3\sqrt{3}}{6\sqrt{3}} = \frac{1}{2},
\frac{BC}{PQ} = \frac{6}{12} = \frac{1}{2}
\]
Hence, $\triangle ABC \sim \triangle PQR$ (by SSS similarity).
Step 2: Corresponding angles of similar triangles
$\triangle ABC \sim \triangle PQR$ means:
\[
\angle A = \angle P, \angle B = \angle Q, \angle C = \angle R
\]
Step 3: Value of $\angle P$
From $\triangle ABC$, $\angle A = 80^\circ$.
So, in $\triangle PQR$, $\angle P = 80^\circ$.
Wait! Let's carefully check the figure again:
- In $\triangle ABC$, $\angle A = 80^\circ$, $\angle B = 60^\circ$.
- Then $\angle C = 180^\circ - (80^\circ+60^\circ) = 40^\circ$.
But from similarity, $\angle C = \angle R$, so $\angle A = \angle P$.
Thus $\angle P = 80^\circ$.
\[
\boxed{\angle P = 80^\circ}
\]
So correct option is (D) $80^\circ$.
In the following figure \(\angle\)MNP = 90\(^\circ\), seg NQ \(\perp\) seg MP, MQ = 9, QP = 4, find NQ. 
Solve the following sub-questions (any four): In \( \triangle ABC \), \( DE \parallel BC \). If \( DB = 5.4 \, \text{cm} \), \( AD = 1.8 \, \text{cm} \), \( EC = 7.2 \, \text{cm} \), then find \( AE \). 
In the following figure, XY \(||\) seg AC. If 2AX = 3BX and XY = 9. Complete the activity to find the value of AC.
Activity:
2AX = 3BX (Given)
\[\therefore \frac{AX}{BX} = \frac{3}{\boxed{2}} \ \\ \frac{AX + BX}{BX} = \frac{3 + 2}{2} \quad \text{(by componendo)} \ \\ \frac{BA}{BX} = \frac{5}{2} \quad \dots \text{(I)} [6pt] \\ \text{Now } \triangle BCA \sim \triangle BYX ; (\boxed{\text{AA}} \text{ test of similarity}) [4pt] \\ \therefore \frac{BA}{BX} = \frac{AC}{XY} \quad \text{(corresponding sides of similar triangles)} [4pt] \\ \frac{5}{2} = \frac{AC}{9} \quad \text{from (I)} [4pt] \\ \therefore AC = \boxed{22.5}\]
Find the unknown frequency if 24 is the median of the following frequency distribution:
\[\begin{array}{|c|c|c|c|c|c|} \hline \text{Class-interval} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\ \hline \text{Frequency} & 5 & 25 & 25 & \text{$p$} & 7 \\ \hline \end{array}\]