Question
✓
Answer
In $\triangle$ABC and $\triangle$PQR,
$\frac{\mathrm{AB}}{\mathrm{RQ}}=\frac{3.8}{7.6}=\frac{1}{2}, \frac{\mathrm{BC}}{\mathrm{QP}}=\frac{6}{12}=\frac{1}{2} $ and $\frac{\mathrm{CA}}{\mathrm{PR}}=\frac{3 \sqrt{3}}{6 \sqrt{3}}=\frac{1}{2}$
That is, $\frac{\mathrm{AB}}{\mathrm{RQ}}=\frac{\mathrm{BC}}{\mathrm{QP}}=\frac{\mathrm{CA}}{\mathrm{PR}}$
So,$\triangle \mathrm{ABC} \sim \triangle \mathrm{RQP}$ (SSS similarity criterion)
Therefore, $\angle$C = $\angle$P (Corresponding angles of similar triangles)
But $\angle$C = 180° – $\angle$A – $\angle$B (Angle sum property of triangle)
$\angle$C = 180° – 80° – 60°
$\angle$C = 40°
So, $\angle$P = 40°
$\frac{\mathrm{AB}}{\mathrm{RQ}}=\frac{3.8}{7.6}=\frac{1}{2}, \frac{\mathrm{BC}}{\mathrm{QP}}=\frac{6}{12}=\frac{1}{2} $ and $\frac{\mathrm{CA}}{\mathrm{PR}}=\frac{3 \sqrt{3}}{6 \sqrt{3}}=\frac{1}{2}$
That is, $\frac{\mathrm{AB}}{\mathrm{RQ}}=\frac{\mathrm{BC}}{\mathrm{QP}}=\frac{\mathrm{CA}}{\mathrm{PR}}$
So,$\triangle \mathrm{ABC} \sim \triangle \mathrm{RQP}$ (SSS similarity criterion)
Therefore, $\angle$C = $\angle$P (Corresponding angles of similar triangles)
But $\angle$C = 180° – $\angle$A – $\angle$B (Angle sum property of triangle)
$\angle$C = 180° – 80° – 60°
$\angle$C = 40°
So, $\angle$P = 40°
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