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 figure, $DE \parallel AC$ and $DF \parallel AE$. Prove that $\dfrac{BF}{FE} = \dfrac{BE}{EC}$.
In the figure $DE \parallel BC$. If $AD = 3\,\text{cm}$, $DE = 4\,\text{cm}$ and $DB = 1.5\,\text{cm}$, then the measure of $BC$ will be:
If \( \triangle ODC \sim \triangle OBA \) and \( \angle BOC = 125^\circ \), then \( \angle DOC = ? \)
Find the unknown frequency if 24 is the median of the following frequency distribution:
\[\begin{array}{|c|c|c|c|c|c|} \hline \text{Class-interval} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\ \hline \text{Frequency} & 5 & 25 & 25 & \text{$p$} & 7 \\ \hline \end{array}\]
Two concentric circles are of radii $8\ \text{cm}$ and $5\ \text{cm}$. Find the length of the chord of the larger circle which touches (is tangent to) the smaller circle.